interference of light waves and young's experiment derivation

Wave Optics

Young’s double slit experiment, learning objectives.

By the end of this section, you will be able to:

• Explain the phenomena of interference.
• Define constructive interference for a double slit and destructive interference for a double slit.

Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see Figure 1).

Figure 1. Young’s double slit experiment. Here pure-wavelength light sent through a pair of vertical slits is diffracted into a pattern on the screen of numerous vertical lines spread out horizontally. Without diffraction and interference, the light would simply make two lines on the screen.

Why do we not ordinarily observe wave behavior for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source (the Sun) through a single slit to make the light somewhat coherent. By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships. Why did Young then pass the light through a double slit? The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single λ ) light to clarify the effect. Figure 2 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

Figure 2. The amplitudes of waves add. (a) Pure constructive interference is obtained when identical waves are in phase. (b) Pure destructive interference occurs when identical waves are exactly out of phase, or shifted by half a wavelength.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 3a. Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 3b. Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

Figure 3. Double slits produce two coherent sources of waves that interfere. (a) Light spreads out (diffracts) from each slit, because the slits are narrow. These waves overlap and interfere constructively (bright lines) and destructively (dark regions). We can only see this if the light falls onto a screen and is scattered into our eyes. (b) Double slit interference pattern for water waves are nearly identical to that for light. Wave action is greatest in regions of constructive interference and least in regions of destructive interference. (c) When light that has passed through double slits falls on a screen, we see a pattern such as this. (credit: PASCO)

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in Figure 4. Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in Figure 4a. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively as shown in Figure 4b. More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [(1/2) λ , (3/2) λ , (5/2) λ , etc.], then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths ( λ , 2 λ , 3 λ , etc.), then constructive interference occurs.

Figure 4. Waves follow different paths from the slits to a common point on a screen. (a) Destructive interference occurs here, because one path is a half wavelength longer than the other. The waves start in phase but arrive out of phase. (b) Constructive interference occurs here because one path is a whole wavelength longer than the other. The waves start out and arrive in phase.

Take-Home Experiment: Using Fingers as Slits

Look at a light, such as a street lamp or incandescent bulb, through the narrow gap between two fingers held close together. What type of pattern do you see? How does it change when you allow the fingers to move a little farther apart? Is it more distinct for a monochromatic source, such as the yellow light from a sodium vapor lamp, than for an incandescent bulb?

Figure 5. The paths from each slit to a common point on the screen differ by an amount dsinθ, assuming the distance to the screen is much greater than the distance between slits (not to scale here).

Figure 5 shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle θ between the path and a line from the slits to the screen (see the figure) is nearly the same for each path. The difference between the paths is shown in the figure; simple trigonometry shows it to be d sin  θ , where d is the distance between the slits. To obtain constructive interference for a double slit , the path length difference must be an integral multiple of the wavelength, or  d sin  θ = mλ, for   m = 0, 1, −1, 2, −2, . . . (constructive).

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

[latex]d\sin\theta=\left(m+\frac{1}{2}\right)\lambda\text{, for }m=0,1,-1,2,-2,\dots\text{ (destructive)}\\[/latex],

where λ is the wavelength of the light, d is the distance between slits, and θ is the angle from the original direction of the beam as discussed above. We call m the order of the interference. For example, m  = 4 is fourth-order interference.

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 6. The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation  d sin  θ = mλ, for   m = 0, 1, −1, 2, −2, . . . .

For fixed λ and m , the smaller d is, the larger θ must be, since [latex]\sin\theta=\frac{m\lambda}{d}\\[/latex]. This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance d apart) is small. Small d gives large θ , hence a large effect.

Figure 6. The interference pattern for a double slit has an intensity that falls off with angle. The photograph shows multiple bright and dark lines, or fringes, formed by light passing through a double slit.

Example 1. Finding a Wavelength from an Interference Pattern

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of 10.95º relative to the incident beam. What is the wavelength of the light?

The third bright line is due to third-order constructive interference, which means that m  = 3. We are given d = 0.0100 mm and θ  = 10.95º. The wavelength can thus be found using the equation  d sin  θ = mλ  for constructive interference.

The equation is d sin  θ = mλ . Solving for the wavelength λ gives [latex]\lambda=\frac{d\sin\theta}{m}\\[/latex].

Substituting known values yields

[latex]\begin{array}{lll}\lambda&=&\frac{\left(0.0100\text{ nm}\right)\left(\sin10.95^{\circ}\right)}{3}\\\text{ }&=&6.33\times10^{-4}\text{ nm}=633\text{ nm}\end{array}\\[/latex]

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with λ , so that spectra (measurements of intensity versus wavelength) can be obtained.

Example 2. Calculating Highest Order Possible

Interference patterns do not have an infinite number of lines, since there is a limit to how big m can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation  d sin  θ = mλ ( for   m = 0, 1, −1, 2, −2, . . . ) describes constructive interference. For fixed values of d and λ , the larger m is, the larger sin  θ  is. However, the maximum value that sin  θ  can have is 1, for an angle of 90º. (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which m corresponds to this maximum diffraction angle.

Solving the equation  d sin  θ = mλ  for m gives [latex]\lambda=\frac{d\sin\theta}{m}\\[/latex].

Taking sin  θ = 1 and substituting the values of d and λ from the preceding example gives

[latex]\displaystyle{m}=\frac{\left(0.0100\text{ mm}\right)\left(1\right)}{633\text{ nm}}\approx15.8\\[/latex]

Therefore, the largest integer m can be is 15, or  m  = 15.

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.

Section Summary

• Young’s double slit experiment gave definitive proof of the wave character of light.
• An interference pattern is obtained by the superposition of light from two slits.
• There is constructive interference when  d sin  θ = mλ ( for   m = 0, 1, −1, 2, −2, . . . ), where d  is the distance between the slits, θ is the angle relative to the incident direction, and m  is the order of the interference.
• There is destructive interference when  d sin  θ = mλ ( for   m = 0, 1, −1, 2, −2, . . . ).

Conceptual Questions

• Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.
• Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.
• Is it possible to create a situation in which there is only destructive interference? Explain.
• Figure 7 shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

Figure 7. This double slit interference pattern also shows signs of single slit interference. (credit: PASCO)

Problems & Exercises

• At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?
• Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.
• What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of 30.0º?
• Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of 45.0º.
• Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00 μm.
• What is the wavelength of light falling on double slits separated by 2.00 μm if the third-order maximum is at an angle of 60.0º?
• At what angle is the fourth-order maximum for the situation in Question 1?
• What is the highest-order maximum for 400-nm light falling on double slits separated by 25.0 μm?
• Find the largest wavelength of light falling on double slits separated by 1.20 μm for which there is a first-order maximum. Is this in the visible part of the spectrum?
• What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?
• (a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?
• (a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of 10.0º, at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?

Figure 8. The distance between adjacent fringes is [latex]\Delta{y}=\frac{x\lambda}{d}\\[/latex], assuming the slit separation d  is large compared with λ .

• Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in Figure 8.
• Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see Figure 8).

coherent:  waves are in phase or have a definite phase relationship

constructive interference for a double slit:  the path length difference must be an integral multiple of the wavelength

destructive interference for a double slit:  the path length difference must be a half-integral multiple of the wavelength

incoherent:  waves have random phase relationships

order:  the integer m  used in the equations for constructive and destructive interference for a double slit

Selected Solutions to Problems & Exercises

3. 1.22 × 10 −6 m

9. 1200 nm (not visible)

11. (a) 760 nm; (b) 1520 nm

13. For small angles sin θ − tan θ  ≈ θ  (in radians).

For two adjacent fringes we have,  d  sin θ m = mλ  and  d sin θ m + 1 = ( m  + 1) λ

Subtracting these equations gives

[latex]\begin{array}{}d\left(\sin{\theta }_{\text{m}+1}-\sin{\theta }_{\text{m}}\right)=\left[\left(m+1\right)-m\right]\lambda \\ d\left({\theta }_{\text{m}+1}-{\theta }_{\text{m}}\right)=\lambda \\ \text{tan}{\theta }_{\text{m}}=\frac{{y}_{\text{m}}}{x}\approx {\theta }_{\text{m}}\Rightarrow d\left(\frac{{y}_{\text{m}+1}}{x}-\frac{{y}_{\text{m}}}{x}\right)=\lambda \\ d\frac{\Delta y}{x}=\lambda \Rightarrow \Delta y=\frac{\mathrm{x\lambda }}{d}\end{array}\\[/latex]

• College Physics. Authored by : OpenStax College. Located at : http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a/College_Physics . License : CC BY: Attribution . License Terms : Located at License

unifyphysics

Young’s Double Slit Experiment

The story of Young’s Double Slit Experiment begins in the early 19th century with a physicist named Thomas Young. At that time, the nature of light was a hotly debated topic. Some scientists, like Isaac Newton, thought light was made up of particles, while others believed it behaved like a wave.

Imagine if light were like tiny balls thrown at a wall with two slits; you’d expect to see two bright spots directly behind the slits on a screen, right? That’s the particle theory. But what if light were more like ripples in a pond, spreading out and overlapping after passing through the slits? This is the wave theory, which suggests a pattern of several bright and dark bands due to the ripples interfering with each other.

In 1801 , Thomas Young set out to settle this debate. He conducted an experiment where he shone light through two closely spaced slits onto a screen. Instead of just two bright spots, he observed a series of alternating bright and dark bands known as interference fringes. This pattern could only be explained if light behaved like waves, overlapping and combining to create the bands.

Young’s experiment was revolutionary because it provided strong evidence for the wave theory of light, which was initially proposed by Christiaan Huygens. It showed that light waves could interfere with each other, just like waves on water, creating patterns of constructive interference (bright bands) where the waves add up, and destructive interference (dark bands) where they cancel out.

Young’s Double Slit Experiment laid the foundation for the field of wave optics and was a precursor to the development of quantum mechanics. It also led to the understanding of the wave-particle duality of light, which is a fundamental concept in modern physics.

By conducting this simple yet ingenious experiment, Thomas Young not only advanced the wave theory of light but also set the stage for future discoveries that would transform our understanding of the universe.

Table of Contents

Interference of Light Waves

Interference is a fundamental phenomenon observed when two or more waves overlap in space, resulting in a new wave pattern. This concept is crucial in understanding various optical effects and is particularly significant in the study of light waves.

When light waves from different sources meet, they don’t just pass by each other. Instead, they interact in such a way that they can add up to make a brighter light (constructive interference) or cancel each other out to create darkness (destructive interference). This interaction is what we call interference.

Coherent Light Sources: For interference to be observable, the light sources need to be coherent. This means they emit waves that have a constant phase difference and the same frequency. Without coherence, the interference pattern would be washed out and not visible.

Constructive Interference: This occurs when the crests (high points) of two waves align together, or the troughs (low points) align. The waves reinforce each other, making a wave that’s bigger than either of the individual waves. This results in a bright spot in the interference pattern.

Destructive Interference: In contrast, destructive interference happens when the crest of one wave meets the trough of another. They effectively cancel each other out, leading to a dark spot in the interference pattern.

Path Difference and Phase Difference: The reason why waves interfere is due to differences in the path they travel (path difference) or their phase (phase difference). When the path difference between two waves is an integer multiple of the wavelength, they are in phase and create constructive interference. When the path difference is a half-integer multiple of the wavelength, they are out of phase and create destructive interference.

Superposition Principle: The principle of superposition states that when two or more waves meet, the resultant wave is the sum of the individual waves. This principle is key to predicting the outcome of interfering light waves.

Understanding interference is essential for explaining many optical phenomena and technologies, such as holography, the iridescence of a soap bubble, and even the working of lasers. It’s a concept that shows the wave-like nature of light and helps us understand how light can behave in such complex and beautiful ways.

Young’s Double Slit Experiment (YDSE) is a classic physics experiment that demonstrates the wave nature of light. It’s named after Thomas Young, who first experimented in 1801 to prove that light can interfere with itself, much like waves do.

In YDSE, light from a single source is split into two separate waves by passing it through two closely spaced slits. When these two waves overlap on the other side, they create an interference pattern of bright and dark fringes on a screen placed behind the slits.

The bright fringes, or bands, occur where the waves from the two slits arrive in phase and reinforce each other, a phenomenon known as constructive interference. The dark fringes appear where the waves are out of phase and cancel each other, which is called destructive interference.

The pattern of fringes can be described mathematically. The equation gives the position of the bright fringes:

\(\displaystyle y_m = \frac{m\lambda D}{d} \)

where (y m ) is the distance from the central fringe to the (m) -th bright fringe, (λ) is the wavelength of the light, (D) is the distance from the slits to the screen, (d) is the distance between the slits, and (m) is the order of the fringe.

YDSE is important because it provides direct evidence of the wave nature of light. Before this experiment, many scientists believed that light consisted of particles. The interference pattern observed in YDSE could not be explained by particle theory, thus supporting the wave theory of light.

Setup: The setup includes a light source, a screen, and a barrier with two closely spaced slits. The light source emits coherent light, which passes through the slits and forms an interference pattern on the screen.

The setup of Young’s Double Slit Experiment is elegantly simple. It consists of:

• A coherent light source, such as a laser, emits light of a single wavelength.
• A barrier with two very narrow, parallel slits separated by a small distance.
• A screen is positioned to capture the light after it passes through the slits.

The light source is placed in such a way that its rays fall directly on the barrier. The slits act as two new sources of light waves, which then spread out due to diffraction.

Working: As light waves pass through the slits, they spread out and overlap, creating regions of constructive interference (bright fringes) and destructive interference (dark fringes).

When the light waves emerging from the two slits meet, they interfere with each other. If the crests of the waves from both slits align, they create a bright spot on the screen due to constructive interference. If a crest from one slit meets a trough from the other, they cancel each other out, resulting in a dark spot due to destructive interference.

This interference pattern of alternating bright and dark fringes is projected onto the screen, demonstrating the wave nature of light. The central bright fringe, known as the central maximum, is the result of the direct overlap of light waves from both slits.

The experiment shows that light waves can be split and combined to form patterns, which is only possible if light behaves as a wave. This setup and the resulting pattern provide a clear demonstration of the fundamental principle of interference in wave optics.

Derivation for YDSE

In Young’s Double Slit Experiment, we use a monochromatic light source (S) placed far from two slits (S 1 ) and (S 2 ). This setup ensures that (S 1 ) and (S 2 ) act as coherent sources, emitting light waves with a constant phase difference. The light waves from these slits interfere with each other on a screen located at a distance (D) from the slits. The distance between two slits is denoted by the letter ‘d.’

Path Difference: When light passes through slits (S 1 ) and (S 2 ) and reaches point ( P ) on the screen, it travels different distances from each slit. This difference in path length creates an interference pattern.

• Path Length from (S 1 ) to (P) : (S 1 P )
• Path Length from (S 2 ) to (P) : (S 2 P )

The path difference (∆) between these two paths is given by:

\(\displaystyle \Delta = S_2P – S_1P \)

For constructive interference (bright fringes), this path difference must be an integer multiple of the wavelength (λ):

\(\displaystyle \Delta = n\lambda \quad \text{(for } n = 0, 1, 2, \ldots) \)

For destructive interference (dark fringes), the path difference must be an odd multiple of half the wavelength:

\(\displaystyle \Delta = \left( n + \frac{1}{2} \right)\lambda \quad \text{(for } n = 0, 1, 2, \ldots) \)

Approximations in Young’s Double Slit Experiment:

Considering the geometry of the setup, if (d) is the distance between the slits and (D) is the distance from the slits to the screen, and assuming (D) is much larger than (d), the path difference (∆) can be approximated using the small-angle approximation.

Approximation 1 : The path difference (∆) can be approximated as:

\(\displaystyle\Delta \approx d \sin \theta \)

Where (θ) is the angle that the line from the midpoint between the slits to point (P) makes with the perpendicular bisector of the line joining the slits.

Approximation 2 : Using the small-angle approximation

(\(\displaystyle \sin \theta \approx \theta \approx \frac{y}{D} \)),

where ( y ) is the vertical displacement from the central point on the screen, we get:

\(\displaystyle \Delta \approx d \frac{y}{D} \)

Position of Fringes

(i) Bright Fringes: Bright fringes, or bright bands, appear on the screen at positions where light from both slits arrives in phase. This means the waves reinforce each other, creating a brighter image. Bright fringes occur where the path difference is an integer multiple of the wavelength.

Substituting the path difference for constructive interference:

\(\displaystyle d \frac{y}{D} = n\lambda \)

Solving for (y):

\(\displaystyle y = \frac{n\lambda D}{d} \)

This equation gives the positions (y) of the bright fringes on the screen.

(ii) Dark Fringes: Dark fringes, or dark bands, occur where light from the slits arrives out of phase, causing the waves to cancel each other out.

For destructive interference, the path difference is:

\(\displaystyle d \frac{y}{D} = \left( n + \frac{1}{2} \right) \lambda \)

Solving for (y)

\(\displaystyle y = \left( n + \frac{1}{2} \right) \frac{\lambda D}{d} \)

This equation gives the positions (y) of the dark fringes on the screen.

(iii) Fringe Width: The fringe width (β) is the distance between two successive bright or dark fringes on the screen. For constructive interference (bright fringes), the position of the (n) -th bright fringe is given by:

\(\displaystyle y_n = \frac{n \lambda D}{d} \)

For the (n+1) -th bright fringe:

\(\displaystyle y_{n+1} = \frac{(n+1) \lambda D}{d} \)

The fringe width (β) is the distance between these two fringes:

\(\displaystyle \beta = y_{n+1} – y_n \) \(\displaystyle \beta = \frac{(n+1) \lambda D}{d} – \frac{n \lambda D}{d} \) \(\displaystyle \beta = \frac{\lambda D}{d} \)

Thus, the fringe width is:

\(\displaystyle \beta = \frac{\lambda D}{d} \)

(iv) Angular Width: The angular width (θ) of a fringe is the angle subtended by the fringe width at the slits.

Since the fringe width is (β) and the distance to the screen is (D), the angular width (∆θ) is:

\(\displaystyle \Delta \theta = \frac{\beta}{D} \)

Substitute the fringe width ( \(\displaystyle\beta = \frac{\lambda D}{d} \)):

\(\displaystyle \Delta \theta = \frac{\frac{\lambda D}{d}}{D}\) \(\displaystyle \Delta \theta = \frac{\lambda}{d} \)

Thus, the angular width of a fringe is:

\(\displaystyle \Delta \theta = \frac{\lambda}{d} \)

(v) Maximum Order: The maximum order of interference fringes is limited by the wavelength and slit separation.

The maximum order (n max ) of the fringe is the highest order bright fringe that can be observed on the screen. For constructive interference, the path difference should be an integer multiple of the wavelength:

\(\displaystyle d \sin \theta = n \lambda \)

The maximum value of ( \(\displaystyle\sin \theta \)) is 1, so:

\(\displaystyle d = n_{\text{max}} \lambda \)

Thus, the maximum order ( \(\displaystyle n_{\text{max}} \)) is:

\(\displaystyle n_{\text{max}} = \frac{d}{\lambda} \)

These expressions help in understanding the spacing and angular spread of the fringes, as well as the maximum number of observable fringes in Young’s Double Slit Experiment.

Shape of Interference Fringes

The fringes are typically hyperbolic but appear straight and parallel due to the large distance between the screen and the slits. In Young’s Double Slit Experiment, the interference pattern that appears on the screen consists of a series of bright and dark bands known as interference fringes. These fringes are the result of constructive and destructive interference between light waves coming from the two slits.

Hyperbolic Nature: The actual shape of these fringes is hyperbolic. This means that if you were to trace the path of the bright and dark bands, they would form hyperbolas. However, because the screen is usually placed far away from the slits (the distance (D) is much larger than the slit separation (d), the fringes appear to be straight and parallel lines.

Why They Appear Straight: The reason the fringes look straight is due to the angles involved. When the screen is far away, the angles at which the light from the two slits meet are very small. This makes the curves of the hyperbolas so gentle that they look like straight lines to the naked eye.

Also Read: Coherent And Incoherent Addition of Waves

Intensity of Fringes

(i) Maximum Intensity: In Young’s Double Slit Experiment, the maximum intensity of fringes refers to the brightest points in the interference pattern.

• The maximum intensity of the fringes occurs at points where the light waves from both slits constructively interfere. This means that the crests of the waves from both slits align perfectly, reinforcing each other and resulting in a bright fringe.
• The central fringe, or the central maximum, is typically the brightest because it is the point of perfect constructive interference, where the path difference between the two waves is zero.

The intensity (I) of the bright fringes can be mathematically represented as proportional to the square of the amplitude of the resultant wave. If (A) is the amplitude of the individual waves, then the intensity at the central maximum is given by:

\(\displaystyle I_{max} = 4A^2 \)

This is because, at the central maximum, the amplitudes of the waves from both slits add up coherently (A + A = 2A), and since intensity is proportional to the square of the amplitude, the intensity becomes (2A) 2 = 4A 2 ).

The concept of maximum intensity is crucial for students to understand because it illustrates the energy distribution across the interference pattern. It shows how the wave nature of light can lead to regions of high energy (brightness) due to the constructive superposition of waves.

(ii) Minimum Intensity: In the context of Young’s Double Slit Experiment, the minimum intensity of fringes refers to the darkest points on the interference pattern.

• The minimum intensity occurs at points of destructive interference, where the waves from the two slits are out of phase.
• At these points, the crest of one wave meets the trough of another, leading to cancellation. The amplitude of the resultant wave is minimized, resulting in a dark fringe.

If the individual waves have an amplitude (A), the amplitude of the resultant wave at the points of destructive interference is (A – A = 0).

Since the intensity of light is proportional to the square of the amplitude, the intensity at these points is (I min = (0) 2 = 0 ).

Practically, due to imperfections in the setup or the quality of the light source, the intensity may not be exactly zero but is significantly lower than that of the bright fringes. These dark fringes are equally spaced between the bright fringes and provide contrast to the pattern, making the bright fringes more noticeable.

Special Cases

(i) Rays Not Parallel to Principal Axis

In Young’s Double Slit Experiment, we usually assume that the light rays coming from the slits to any point on the screen are parallel. This assumption simplifies the analysis and is valid when the screen is far away from the slits. However, there are special cases where this assumption does not hold, particularly when the rays are not parallel.

When the rays from the slits are not parallel, it means they are converging or diverging as they travel toward the screen. This can happen if the light source is not placed far enough from the slits or if the slits themselves are not aligned properly.

Non-parallel rays can cause the interference pattern to shift or distort. Instead of a symmetrical pattern centered on the normal (the line perpendicular to the screen and passing through the midpoint between the slits), the fringes may be skewed or displaced.

To understand this concept, imagine the light waves as water ripples in a pond. If you drop two stones at different distances from the shore, the ripples won’t meet at the same angle along the entire shoreline. Similarly, if the light waves from the slits are not parallel, they will interfere at different angles, affecting the symmetry of the fringe pattern.

When rays are not parallel to the principal axis, we need to consider the angle at which the light source is positioned relative to the central line between the two slits.

When rays are not parallel to the principal axis, the path difference calculation changes due to the different angles of incidence.

• (θ) is the angle of incidence.
• (∆x ) is the path difference.

Path Difference :

\(\displaystyle\Delta x = (AS_1 + S_1P) – S_2P\)

From the geometry:

\(\displaystyle\Delta x = AS_1 – (S_2P – S_1P)\)

\(\displaystyle\Delta x = d \sin \theta – \frac{4d}{D}y\)

Condition for Maxima and Minima :

For maxima:

\(\displaystyle\Delta x = n\lambda\)

For minima:

\(\displaystyle\Delta x = (2n – 1) \frac{\lambda}{2}\)

Using these equations, we can calculate different positions of maxima and minima.

(ii) Source Placed Beyond the Central Line

In Young’s Double Slit Experiment, a special case arises when the light source is placed beyond the central line that bisects the distance between the two slits. Typically, the light source in Young’s Double Slit Experiment is aligned such that it is equidistant from both slits. This ensures that the light waves originating from the slits have symmetrical paths to any point on the screen, resulting in a symmetrical interference pattern.

If the source is placed beyond the central line, the symmetry is broken. One slit is now closer to the source than the other. This means that the light waves from the closer slit will have a shorter path to travel to reach any point on the screen compared to waves from the farther slit.

When the source is placed above or below the central line, the path difference calculation changes due to the different distances to the slits.

• (S 1 ) and (S 2 ) are the slits.
• (P) is a point on the screen.
• (d) is the distance between the slits.
• (D) is the distance from the slits to the screen.
• (a) is the horizontal distance between the source and the central line.
• (b) is the vertical distance between the source and the slits.
• (Δx) is the path difference.

\(\displaystyle\Delta x = (\text{distance of ray 2}) – (\text{distance of ray 1})\)

\(\displaystyle\Delta x = (S_2P + S_2P) – (S_1S + S_1P)\) \(\displaystyle\Delta x = (S_2P + S_1) – (S_2P – S_1P)\) \(\displaystyle\Delta x = \frac{bd}{a} + \frac{yd}{D}\)

By knowing the value of (∆x ) from the above expressions, we can calculate different positions of maxima and minima.

The result is a shift in the interference pattern. The central maximum may no longer be directly in front of the midpoint between the two slits. Instead, it shifts towards the slit closer to the source. The entire pattern of bright and dark fringes will also shift accordingly, and the fringes may no longer be evenly spaced across the screen.

This shift can be understood by considering the path difference between the waves from the two slits. Since the path difference changes due to the off-center placement of the source, the conditions for constructive and destructive interference at various points on the screen also change.

Displacement of Fringes

In Young’s Double Slit Experiment, the displacement of fringes refers to the shift in the position of the bright and dark bands on the screen due to changes in the experimental setup or external conditions.

Causes of Displacement:

• Change in Wavelength : If the wavelength of light used in the experiment changes, the fringe pattern will shift. A longer wavelength will cause the fringes to move further apart, while a shorter wavelength will bring them closer together.
• Change in Medium : Introducing a different medium (like water or glass) into the path of one of the beams can alter the speed of light in that path, leading to a shift in the interference pattern.
• Movement of the Screen or Slits : Physically moving the screen or the slits closer or further away from each other can also cause the fringes to shift.

The displacement can be quantified by measuring the change in position of the central bright fringe or any other fringe. This shift is directly related to the path difference between the light waves from the two slits.

When a thin transparent plate of thickness (t) is introduced in front of one of the slits in Young’s double-slit experiment, the fringe pattern shifts toward the side where the plate is present. This is due to the change in optical path length caused by the introduction of the plate. Let’s derive the expression for the displacement of fringes.

• (S 1 ) and (S 2 ) are the slits separated by distance (d).
• (P) and (P 1 ) are points on the screen where fringes are observed.
• ( S ) is a point on the screen where a fringe is observed without the plate.
• A transparent plate of thickness (t) and refractive index (µ) is placed in front of slit (S 2 ).

Path Difference Before Introducing the Plate

Without the Plate: The path difference between the light rays from (S 1 ) and (S 2 ) at point (P) is:

\(\displaystyle\delta = S_1P – S_2P\)

Path Difference After Introducing the Plate

With the Plate:

The path length for the light ray passing through (S 2 ) changes due to the plate. The new optical path length for the light from (S 2 ) is:

\(\displaystyle S_2P_1 = (S_2P_1 – t){\text{air}} + t{\text{plate}} = (S_2P_1 – t) + \mu t\)

Simplifying this, we get:

\(\displaystyle S_2P_1 = S_2P_1 + (\mu – 1)t\)

New Path Difference:

The new path difference between the light rays from (S 1 ) and (S 2 ) at point (P 1 ) is:

\(\displaystyle(\delta){\text{new}} = S_1P_1 – S_2P_1 = S_1P_1 – (S_2P_1 + (\mu – 1)t) \)

\(\displaystyle (\delta x){\text{new}} = (S_1P_1 – S_2P_1) – (\mu – 1)t\)

Using the original path difference ( \(\displaystyle\delta x = d \sin \theta\) ):

\(\displaystyle (\delta x)_{\text{new}} = d \sin \theta – (\mu – 1)t\)

Fringe Shift Calculation

Fringe Position: The position of fringes is given by:

\(\displaystyle y = \frac{\delta x D}{d}\)

For the new path difference:

\(\displaystyle y_{\text{new}} = \frac{(\Delta x)_{\text{new}} D}{d}\)

\(\displaystyle = \frac{(d \sin \theta – (\mu – 1)t) D}{d}\)

Simplifying, we get:

\(\displaystyle y_{\text{new}} = \frac{d \sin \theta D}{d} – \frac{(\mu – 1)t D}{d}\)

\(\displaystyle y_{\text{new}} = D \sin \theta – \frac{(\mu – 1)t D}{d}\)

Fringe Displacement: The displacement of the fringes (\(\displaystyle \delta y\)) is the difference between the new and the original fringe positions:

\(\displaystyle\delta y = y_{\text{new}} – y\)

\(\displaystyle\delta y = \left( D \sin \theta – \frac{(\mu – 1)t D}{d} \right) – D \sin \theta\)

\(\displaystyle\delta y = – \frac{(\mu – 1)t D}{d}\)

Thus, the expression for the displacement of fringes in Young’s double-slit experiment when a thin transparent plate is introduced is:

\(\displaystyle \delta y = – \frac{(\mu – 1)t D}{d} \)

This shows that the fringe pattern shifts towards the side where the plate is present by an amount (\(\displaystyle\frac{(\mu – 1)t D}{d}\)).

Constructive and Destructive Interference

Constructive Interference :

What is Constructive Interference? – Constructive interference occurs when two or more waves meet and their displacements are in the same direction. This means that the crests (the highest points) of one wave align with the crests of another wave, and the troughs (the lowest points) align as well.

When waves interfere constructively, their amplitudes add together. The amplitude is the height of the wave from the center line to the crest. So, if two waves with the same amplitude interfere constructively, the resultant wave will have an amplitude that is double the individual waves’ amplitudes.

For constructive interference to occur, the waves must have a phase difference that is an even multiple of π (180°). This means the waves are ‘in phase’ and their peaks and troughs match up perfectly as they combine.

Imagine two ripples in a pond that are heading towards each other. If they meet and the peak of one ripple meets the peak of the other, they will combine to create a higher ripple. This is constructive interference in action.

Destructive Interference :

Occurs when the crest of one wave aligns with the trough of another, resulting in darkness or cancellation. When two waves of the same frequency and wavelength meet, if the crest of one wave aligns with the trough of another, they interfere destructively. This happens because the high point of one wave fills in the low point of the other, resulting in a wave with a reduced (or zero) amplitude.

The waves must be out of phase by an odd multiple of half the wavelength, which means the path difference between them is

\(\displaystyle (2n-1)\frac{\lambda}{2} \),

where (n) is an integer, and (λ) is the wavelength.

The phase difference between the waves should be (\(\displaystyle (2n-1)\pi \)), indicating that they are half a cycle out of sync.

The amplitude of the resulting wave at points of destructive interference is significantly lower than the individual waves. If the waves have equal amplitude, the resulting amplitude can be zero, leading to a point of no wave movement, known as a node.

Sample Questions

Problem: Derive the expression for the fringe spacing (fringe width) in Young’s Double Slit Experiment.

Solution: In YDSE, light from a coherent source is split into two waves by two slits, (S 1 ) and (S 2 ), separated by a distance (d). These waves interfere with a screen placed at a distance (D) from the slits.

Path Difference: The path difference (∆x ) between the two waves arriving at a point (P) on the screen is given by:

\(\displaystyle\Delta x = d \sin \theta\)

For small angles, ( \(\displaystyle\sin \theta \approx \tan \theta = \frac{y}{D} \)), where (y) is the distance from the central maximum to point (P):

\(\displaystyle\Delta x = d \frac{y}{D}\)

Constructive Interference: For constructive interference (bright fringes), the path difference must be an integer multiple of the wavelength (λ):

\(\displaystyle d \frac{y_n}{D} = n\lambda \quad \Rightarrow \quad y_n = \frac{n\lambda D}{d}\)

Fringe Spacing: The fringe spacing (or fringe width) (β) is the distance between two successive bright fringes:

\(\displaystyle\beta = y_{n+1} – y_n = \frac{(n+1)\lambda D}{d} – \frac{n\lambda D}{d} = \frac{\lambda D}{d}\)

The fringe spacing (fringe width) in Young’s Double Slit Experiment is given by (\(\displaystyle \beta = \frac{\lambda D}{d}\) ).

Problem: Derive the expression for the intensity at any point on the screen in Young’s Double Slit Experiment.

Solution: The intensity at a point on the screen in YDSE depends on the superposition of the light waves from the two slits.

Electric Fields: Let the electric fields due to the slits (S 1 ) and (S 2 ) be (E 1 ) and (E 2 ), respectively. Assuming equal amplitudes (E 0 ):

\(\displaystyle E_1 = E_0 \cos (\omega t)\) \(\displaystyle E_2 = E_0 \cos (\omega t + \phi)\)

where (\(\displaystyle \phi \)) is the phase difference due to the path difference (∆x).

Resultant Electric Field: The resultant electric field (E) is:

\(\displaystyle E = E_1 + E_2 = E_0 \cos (\omega t) + E_0 \cos (\omega t + \phi)\)

Using the trigonometric identity for the sum of cosines:

\(\displaystyle E = 2E_0 \cos \left( \frac{\phi}{2} \right) \cos \left( \omega t + \frac{\phi}{2} \right)\)

Intensity: The intensity (I) is proportional to the square of the amplitude of the resultant electric field:

\(\displaystyle I = E^2 \propto \left[ 2E_0 \cos \left( \frac{\phi}{2} \right) \right]^2 = 4E_0^2 \cos^2 \left( \frac{\phi}{2} \right)\)

Let (I 0 ) be the intensity due to one slit. Therefore, the total intensity (I) is:

\(\displaystyle I = 4I_0 \cos^2 \left( \frac{\phi}{2} \right)\)

Phase Difference: The phase difference (\(\displaystyle \phi\) ) is related to the path difference (∆x) by (\(\displaystyle \phi = \frac{2\pi}{\lambda} \Delta x\) ). For small angles, (\(\displaystyle\Delta x = d \sin \theta \approx d \frac{y}{D}\)):

\(\displaystyle\phi = \frac{2\pi}{\lambda} d \frac{y}{D}\)

Thus, the intensity at a point (y) on the screen is:

\(\displaystyle I = 4I_0 \cos^2 \left( \frac{\pi d y}{\lambda D} \right)\)

The intensity at a point (y) on the screen in Young’s Double Slit Experiment is given by (\(\displaystyle I = 4I_0 \cos^2 \left( \frac{\pi d y}{\lambda D} \right) \)), where (I 0 ) is the intensity due to one slit.

Problem: Calculate the position of the third bright fringe on a screen 2 m away from the slits in Young’s Double Slit Experiment, given (d = 0.1) mm and (λ = 600) nm.

Solution: The position of the ( n ) -th bright fringe is given by:

\(\displaystyle y_n = \frac{n\lambda D}{d}\)

For the third bright fringe (n = 3):

\(\displaystyle y_3 = \frac{3 \times 600 \times 10^{-9} \times 2}{0.1 \times 10^{-3}}\)

\(\displaystyle y_3 = \frac{3 \times 600 \times 10^{-9} \times 2}{0.1 \times 10^{-3}} = \frac{3 \times 600 \times 2}{0.1} \times 10^{-6} \)

\(\displaystyle = \frac{3600}{0.1} \times 10^{-6} = 36000 \times 10^{-6} = 0.036 \text{ m}\)

The position of the third bright fringe is 0.036 m (36 mm) from the central maximum.

Problem: A glass plate of thickness 0.5 mm and refractive index 1.5 is introduced in the path of one of the beams in Young’s Double Slit Experiment. Calculate the shift in the central fringe.

Solution: The introduction of a glass plate changes the optical path length of the beam passing through it. The optical path length change (∆L) is given by:

\(\displaystyle\Delta L = (n – 1)t\)

Substitute the given values:

\(\displaystyle\Delta L = (1.5 – 1) \times 0.5 \times 10^{-3} = 0.5 \times 0.5 \times 10^{-3} = 0.25 \times 10^{-3} \text{ m} = 250 \text{ nm}\)

The path difference (∆x ) introduced by the plate causes a shift in the fringe pattern. The fringe shift (∆y ) can be calculated using the fringe spacing formula:

\(\displaystyle\Delta y = \frac{\Delta L \cdot D}{d}\)

Assuming (D = 2 m) and (d = 0.1 mm = 0.1 × 10 -3 m):

\(\displaystyle\Delta y = \frac{250 \times 10^{-9} \times 2}{0.1 \times 10^{-3}} = \frac{250 \times 2}{0.1} \times 10^{-6} \)

\(\displaystyle = \frac{500}{0.1} \times 10^{-6} = 5000 \times 10^{-6} = 0.005 \text{ m} = 5 \text{ mm}\)

The introduction of a glass plate of thickness 0.5 mm and refractive index 1.5 in the path of one of the beams in YDSE causes the central fringe to shift by 5 mm.

What is Young’s Double Slit Experiment (YDSE) and why is it significant?

Young’s Double Slit Experiment is a foundational experiment in physics that demonstrates the wave nature of light. By allowing light to pass through two closely spaced slits and observing the resulting interference pattern on a screen, the experiment provides evidence for the principle of superposition and the wave behavior of light. It is significant because it challenged the classical particle view of light and supported the wave theory.

What is the basic derivation for the fringe pattern in YDSE?

The basic derivation for the fringe pattern in YDSE involves considering the path difference between light waves coming from the two slits to a point on the screen. This path difference leads to constructive or destructive interference, creating bright and dark fringes. The positions of these fringes can be derived by analyzing the geometry of the setup and applying the principle of superposition.

What approximations are commonly used in Young’s Double Slit Experiment?

Common approximations in YDSE include assuming the screen is far from the slits (far-field approximation) so that the light rays are nearly parallel, and considering the slit separation to be much smaller than the distance to the screen. These approximations simplify the calculations and allow for the use of linear approximations for the path difference.

How is the position of the fringes determined in YDSE?

The position of the fringes in YDSE is determined by the condition for constructive and destructive interference. For constructive interference (bright fringes), the path difference between the waves from the two slits must be an integer multiple of the wavelength. For destructive interference (dark fringes), the path difference must be an odd multiple of half wavelengths. These conditions give the positions of the bright and dark fringes on the screen.

How is the intensity of fringes calculated in YDSE?

The intensity of fringes in YDSE is calculated based on the principle of superposition of the electric fields from the two slits. Constructive interference results in maximum intensity, while destructive interference results in minimum (or zero) intensity. The intensity pattern follows a cosine-squared distribution for constructive and destructive interference points.

What happens if the rays in YDSE are not parallel?

If the rays in YDSE are not parallel, the interference pattern can become distorted. The assumption of parallel rays is an approximation that simplifies the analysis. In reality, if the rays diverge or converge, the fringe spacing may change, and the fringes may not be perfectly straight or evenly spaced. This deviation can be accounted for with more complex calculations that do not assume parallel rays.

How does placing the source beyond the central line affect the fringe pattern and what is fringe displacement?

Placing the source beyond the central line in YDSE causes the entire fringe pattern to shift sideways. This shift is known as fringe displacement. The central maximum (the bright fringe at the center) moves from the geometric center of the screen to a new position corresponding to the new optical path difference. This displacement can be calculated and is used in various applications to measure small changes in optical path lengths or displacements.

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The observation of interference effects definitively indicates the presence of overlapping waves. Thomas Young postulated that light is a wave and is subject to the superposition principle; his great experimental achievement was to demonstrate the constructive and destructive interference of light (c. 1801). In a modern version of Young’s experiment, differing in its essentials only in the source of light, a laser equally illuminates two parallel slits in an otherwise opaque surface. The light passing through the two slits is observed on a distant screen. When the widths of the slits are significantly greater than the wavelength of the light, the rules of geometrical optics hold—the light casts two shadows, and there are two illuminated regions on the screen. However, as the slits are narrowed in width, the light diffracts into the geometrical shadow, and the light waves overlap on the screen. (Diffraction is itself caused by the wave nature of light, being another example of an interference effect—it is discussed in more detail below.)

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The superposition principle determines the resulting intensity pattern on the illuminated screen. Constructive interference occurs whenever the difference in paths from the two slits to a point on the screen equals an integral number of wavelengths (0, λ, 2λ,…). This path difference guarantees that crests from the two waves arrive simultaneously. Destructive interference arises from path differences that equal a half-integral number of wavelengths (λ/2, 3λ/2,…). Young used geometrical arguments to show that the superposition of the two waves results in a series of equally spaced bands, or fringes, of high intensity, corresponding to regions of constructive interference, separated by dark regions of complete destructive interference.

An important parameter in the double-slit geometry is the ratio of the wavelength of the light λ to the spacing of the slits d . If λ/ d is much smaller than 1, the spacing between consecutive interference fringes will be small, and the interference effects may not be observable. Using narrowly separated slits, Young was able to separate the interference fringes. In this way he determined the wavelengths of the colours of visible light. The very short wavelengths of visible light explain why interference effects are observed only in special circumstances—the spacing between the sources of the interfering light waves must be very small to separate regions of constructive and destructive interference.

Observing interference effects is challenging because of two other difficulties. Most light sources emit a continuous range of wavelengths, which result in many overlapping interference patterns, each with a different fringe spacing. The multiple interference patterns wash out the most pronounced interference effects, such as the regions of complete darkness. Second, for an interference pattern to be observable over any extended period of time, the two sources of light must be coherent with respect to each other. This means that the light sources must maintain a constant phase relationship. For example, two harmonic waves of the same frequency always have a fixed phase relationship at every point in space, being either in phase, out of phase, or in some intermediate relationship. However, most light sources do not emit true harmonic waves; instead, they emit waves that undergo random phase changes millions of times per second. Such light is called incoherent . Interference still occurs when light waves from two incoherent sources overlap in space, but the interference pattern fluctuates randomly as the phases of the waves shift randomly. Detectors of light, including the eye, cannot register the quickly shifting interference patterns, and only a time-averaged intensity is observed. Laser light is approximately monochromatic (consisting of a single wavelength) and is highly coherent; it is thus an ideal source for revealing interference effects.

After 1802, Young’s measurements of the wavelengths of visible light could be combined with the relatively crude determinations of the speed of light available at the time in order to calculate the approximate frequencies of light. For example, the frequency of green light is about 6 × 10 14 Hz ( hertz , or cycles per second). This frequency is many orders of magnitude larger than the frequencies of common mechanical waves. For comparison, humans can hear sound waves with frequencies up to about 2 × 10 4 Hz. Exactly what was oscillating at such a high rate remained a mystery for another 60 years.

Chapter 27 Wave Optics

27.3 Young’s Double Slit Experiment

• Explain the phenomena of interference.
• Define constructive interference for a double slit and destructive interference for a double slit.

Although Christiaan Huygens thought that light was a wave, Isaac Newton did not. Newton felt that there were other explanations for color, and for the interference and diffraction effects that were observable at the time. Owing to Newton’s tremendous stature, his view generally prevailed. The fact that Huygens’s principle worked was not considered evidence that was direct enough to prove that light is a wave. The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773–1829) did his now-classic double slit experiment (see Figure 1 ).

Why do we not ordinarily observe wave behavior for light, such as observed in Young’s double slit experiment? First, light must interact with something small, such as the closely spaced slits used by Young, to show pronounced wave effects. Furthermore, Young first passed light from a single source (the Sun) through a single slit to make the light somewhat coherent. By coherent , we mean waves are in phase or have a definite phase relationship. Incoherent means the waves have random phase relationships. Why did Young then pass the light through a double slit? The answer to this question is that two slits provide two coherent light sources that then interfere constructively or destructively. Young used sunlight, where each wavelength forms its own pattern, making the effect more difficult to see. We illustrate the double slit experiment with monochromatic (single [latex]{\lambda}[/latex]) light to clarify the effect. Figure 2 shows the pure constructive and destructive interference of two waves having the same wavelength and amplitude.

When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 3 (a). Pure constructive interference occurs where the waves are crest to crest or trough to trough. Pure destructive interference occurs where they are crest to trough. The light must fall on a screen and be scattered into our eyes for us to see the pattern. An analogous pattern for water waves is shown in Figure 3 (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. These angles depend on wavelength and the distance between the slits, as we shall see below.

To understand the double slit interference pattern, we consider how two waves travel from the slits to the screen, as illustrated in Figure 4 . Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they may end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively as shown in Figure 4 (a). If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively as shown in Figure 4 (b). More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [[latex]{(1/2) \;\lambda}[/latex], [latex]{(3/2) \;\lambda}[/latex], [latex]{(5/2) \;\lambda}[/latex], etc.], then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths ([latex]{\lambda}[/latex], [latex]{2 \lambda}[/latex], [latex]{3 \lambda}[/latex], etc.), then constructive interference occurs.

Take-Home Experiment: Using Fingers as Slits

Look at a light, such as a street lamp or incandescent bulb, through the narrow gap between two fingers held close together. What type of pattern do you see? How does it change when you allow the fingers to move a little farther apart? Is it more distinct for a monochromatic source, such as the yellow light from a sodium vapor lamp, than for an incandescent bulb?

Figure 5 shows how to determine the path length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle [latex]{\theta}[/latex] between the path and a line from the slits to the screen (see the figure) is nearly the same for each path. The difference between the paths is shown in the figure; simple trigonometry shows it to be [latex]{d \;\text{sin} \;\theta}[/latex], where [latex]{d}[/latex] is the distance between the slits. To obtain constructive interference for a double slit , the path length difference must be an integral multiple of the wavelength, or

Similarly, to obtain destructive interference for a double slit , the path length difference must be a half-integral multiple of the wavelength, or

where [latex]{\lambda}[/latex] is the wavelength of the light, [latex]{d}[/latex] is the distance between slits, and [latex]{\theta}[/latex] is the angle from the original direction of the beam as discussed above. We call [latex]{m}[/latex] the order of the interference. For example, [latex]{m = 4}[/latex] is fourth-order interference.

The equations for double slit interference imply that a series of bright and dark lines are formed. For vertical slits, the light spreads out horizontally on either side of the incident beam into a pattern called interference fringes, illustrated in Figure 6 . The intensity of the bright fringes falls off on either side, being brightest at the center. The closer the slits are, the more is the spreading of the bright fringes. We can see this by examining the equation

For fixed [latex]{\lambda}[/latex] and [latex]{m}[/latex], the smaller [latex]{d}[/latex] is, the larger [latex]{\theta}[/latex] must be, since [latex]{\text{sin} \;\theta = m \lambda / d}[/latex].

This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance [latex]{d}[/latex] apart) is small. Small [latex]{d}[/latex] gives large [latex]{\theta}[/latex], hence a large effect.

Example 1: Finding a Wavelength from an Interference Pattern

Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find that the third bright line on a screen is formed at an angle of [latex]{10.95 ^{\circ}}[/latex] relative to the incident beam. What is the wavelength of the light?

The third bright line is due to third-order constructive interference, which means that [latex]{m = 3}[/latex]. We are given [latex]{d = 0.0100 \;\text{mm}}[/latex] and [latex]{\theta = 10.95^{\circ}}[/latex]. The wavelength can thus be found using the equation [latex]{d \;\text{sin} \;\theta = m \lambda}[/latex] for constructive interference.

The equation is [latex]{d \;\text{sin} \;\theta = m \lambda}[/latex]. Solving for the wavelength [latex]{\lambda}[/latex] gives

Substituting known values yields

To three digits, this is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did this for visible wavelengths. This analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with [latex]{\lambda}[/latex], so that spectra (measurements of intensity versus wavelength) can be obtained.

Example 2: Calculating Highest Order Possible

Interference patterns do not have an infinite number of lines, since there is a limit to how big [latex]{m}[/latex] can be. What is the highest-order constructive interference possible with the system described in the preceding example?

Strategy and Concept

The equation [latex]{d \;\text{sin} \;\theta = m \lambda \; (\text{for} \; m = 0, \; 1, \; -1, \; 2, \; -2, \; \dots)}[/latex] describes constructive interference. For fixed values of [latex]{d}[/latex] and [latex]{\lambda}[/latex], the larger [latex]{m}[/latex] is, the larger [latex]{\text{sin} \;\theta}[/latex] is. However, the maximum value that [latex]{\text{sin} \;\theta}[/latex] can have is 1, for an angle of [latex]{90 ^{\circ}}[/latex]. (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which [latex]{m}[/latex] corresponds to this maximum diffraction angle.

Solving the equation [latex]{d \;\text{sin} \;\theta = m \lambda}[/latex] for [latex]{m}[/latex] gives

Taking [latex]{\text{sin} \;\theta = 1}[/latex] and substituting the values of [latex]{d}[/latex] and [latex]{\lambda}[/latex] from the preceding example gives

Therefore, the largest integer [latex]{m}[/latex] can be is 15, or

The number of fringes depends on the wavelength and slit separation. The number of fringes will be very large for large slit separations. However, if the slit separation becomes much greater than the wavelength, the intensity of the interference pattern changes so that the screen has two bright lines cast by the slits, as expected when light behaves like a ray. We also note that the fringes get fainter further away from the center. Consequently, not all 15 fringes may be observable.

Section Summary

• Young’s double slit experiment gave definitive proof of the wave character of light.
• An interference pattern is obtained by the superposition of light from two slits.
• There is constructive interference when [latex]{d \;\text{sin} \;\theta = m \lambda \;(\text{for} \; m = 0, \; 1, \; -1, \;2, \; -2, \dots)}[/latex], where [latex]{d}[/latex] is the distance between the slits, [latex]{\theta}[/latex] is the angle relative to the incident direction, and [latex]{m}[/latex] is the order of the interference.
• There is destructive interference when [latex]{d \;\text{sin} \;\theta = (m+ \frac{1}{2}) \lambda}[/latex] (for [latex]{m = 0, \; 1, \; -1, \; 2, \; -2, \; \dots}[/latex]).

Conceptual Questions

1: Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.

2: Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.

3: Is it possible to create a situation in which there is only destructive interference? Explain.

4:   Figure 7 shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.

Problems & Exercises

2: Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.

3: What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of [latex]{30.0 ^{\circ}}[/latex]?

4: Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of [latex]{45.0 ^{\circ}}[/latex].

5: Calculate the wavelength of light that has its third minimum at an angle of [latex]{30.0 ^{\circ}}[/latex] when falling on double slits separated by [latex]{3.00 \;\mu \text{m}}[/latex]. Explicitly, show how you follow the steps in Chapter 27.7 Problem-Solving Strategies for Wave Optics .

6: What is the wavelength of light falling on double slits separated by [latex]{2.00 \;\mu \text{m}}[/latex] if the third-order maximum is at an angle of [latex]{60.0 ^{\circ}}[/latex]?

7: At what angle is the fourth-order maximum for the situation in Problems & Exercises 1 ?

8: What is the highest-order maximum for 400-nm light falling on double slits separated by [latex]{25.0 \;\mu \text{m}}[/latex]?

9: Find the largest wavelength of light falling on double slits separated by [latex]{1.20 \;\mu \text{m}}[/latex] for which there is a first-order maximum. Is this in the visible part of the spectrum?

10: What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?

11: (a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light? (b) For all visible light?

12: (a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of [latex]{10.0^{\circ}}[/latex], at what angle is the second-order maximum? (b) What is the angle of the first minimum? (c) What is the highest-order maximum possible here?

13:   Figure 8 shows a double slit located a distance [latex]{x}[/latex] from a screen, with the distance from the center of the screen given by [latex]{y}[/latex]. When the distance [latex]{d}[/latex] between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, for small angles (where [latex]{\text{sin} \;\theta \approx \theta}[/latex], with [latex]{\theta}[/latex] in radians), the distance between fringes is given by [latex]{\Delta y = x \lambda /d}[/latex].

14: Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in Figure 8 .

15: Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see Figure 8 ).

1:  [latex]{0.516 ^{\circ}}[/latex]

3:  [latex]{1.22 \times 10^{-6} \;\text{m}}[/latex]

7: [latex]{2.06 ^{\circ}}[/latex]

9: 1200 nm (not visible)

11: (a) 760 nm

(b) 1520 nm

13: For small angles [latex]{\text{sin} \;\theta - \;\text{tan} \;\theta \approx \theta}[/latex] (in radians).

For two adjacent fringes we have,

Subtracting these equations gives

College Physics Copyright © August 22, 2016 by OpenStax is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Young’s double slit experiment derivation

One of the first demonstration of the intererference of light waves was given by Young – an English physicist in 1801. We have learnt that two essential conditions to obtain an interference phenomenon are :

• Two sources should be coherent and
• Two coherent sources must be placed close to each other as the wavelength of light is very small.
• 1 Young’s double slit experiment derivation
• 2 Theory of the Experiment
• 3.1 Maxima or Bright fringes
• 3.2 Minima or Dark fringes
• 4.1 Double slit experiment formula?
• 4.2 Fringe width formula in Young’s experiment?

Young placed a monochromatic source (S) of light in front of a narrow slit S 0 and arranged two very narrow slits S₁ and S₂ close to each other in front of slit S 0 young’s double slit experiment derivation diagram below. Slits S₁ and S₂ are equidistant from S 0 , so the spherical wavefronts emitted by slit S 0 reach the slits S₁ and S₂ in equal time.

These wavefronts after arriving at S₁ and S₂ spread out of these slits. Thus the emerging waves are of the same amplitude and wavelength and are in phase. Hence slits S₁ and S₂ behave as coherent sources.

The wavefronts emitted by coherent sources S₁ and S₂ superpose and give rise to interference . When these wavefronts are received on the screen, interference fringes are seen as shown in young’s double slit experiment diagram below.

The points where the destructive interference takes place, we get minima or dark fringe and where the constructive interference takes place, maxima or bright fringe is obtained. The pattern of these dark and bright fringes obtained on the screen is called interference pattern.

Young had used sun light as source of light and circular slits in his experiment.

Theory of the Experiment

Suppose S is the monochromatic source of light. S 0 is the slit through which the light passes and illuminates the slits S₁ and S₂. The waves emitted by slits S₁ and S₂ are the part of the same wavefront, so these waves have the same frequency and the same phase.

Hence slits S 1 and S 2 behave as two coherent sources. Interference takes place on the screen. If we consider a point O on the perpendicular bisector of S₁S 2 , the waves traveling along S₁O and S₂O have traveled equal distances. Hence they will arrive at O in phase and interfere constructively to make O the centre of a bright fringe or maxima.

Derivation of Young’s double slit experiment

To locate the position of the maxima and minima on both sides of O, consider any point P at a distance x from O. Join S 1 P and S 2 P. Now draw S 1 N normal on S 2 P. Then the path difference between S 2 P and S 1 P

Now from △ S 1 PL,

and from △ S 2 PM,

Since the distance of screen from slits S 1 and S 2 is very large, so S 2 P ≈S 1 P ≈D

Path difference,

Maxima or Bright fringes

If the path difference (S 2 P-S 1 P) = xd/D is an integral multiple of λ, then the point P will be the position of bright fringe or maxima.

That is for bright fringe,

Eqn. (1) gives the position of different bright fringes.

P = 0, x =0 , i.e., the central fringe at O will be bright.

This is the position of first bright fringe w.r.t. point O.

This is the position of second bright fringe w.r.t. point O.

………………………………………………………………………………………

This is the position of pth bright fringe w.r.t. point O.

This is the position of (p+1) bright fringe w.r.t. point O.

The distance between two successive bright fringes is called fringe width and is given by

Minima or Dark fringes

If the path difference (S 2 P-S 1 P)=xd/D is an odd multiple of λ/2 , then the point P will be the position of dark fringes or minima.

Thus for dark fringes,

Eqn. (3) gives the position of different dark fringes.

This is the position of first dark fringe w.r.t. point O.

This is the position of second dark fringe w.r.t. point O.

This is the position of third dark fringe w.r.t. point O.

…………………………………………………………………………….

This is the position of pth dark fringe w.r.t. point O.

This is the position of (p+1) dark fringe w.r.t. point O.

The distance between two successive dark fringes is called fringe width (β) of the dark fringes which is given by

This eqn. (4), ‘ β = λD/d’ is called Fringe width formula in Young’s experiment .

From eqns. (2) and (4), it is evident that the fringes width of bright fringe and dark fringe is the same.

If we know the value of “D” and “d” then the measurement of the fringe width ( β ) gives a direct determination of the wavelength of light.

FAQ on Young’s double slit experiment derivation

Double slit experiment formula.

In a double-slit experiment,   λ= xd / L is the formula for the calculation of wavelength.

Fringe width formula in Young’s experiment?

If we know the value of “D” and “d” then the measurement of the fringe width ( β ) gives a direct determination of the wavelength of light. Fringe width formula in Young’s experiment is given by: β = λD/d

Young's Double Slit Interference

Wave Optics Copyright © 1996, 1997 Sergey G. Vtorov. All rights reserved.

Young's Double-Slit Experiment ( AQA A Level Physics ) Revision note. Double Slit Interference The interference of two coherent wave sources  A single wave source passing through a double slit The laser light source is placed behind the single slit So the light is diffracted, producing two light sources at slits A  and B The light from the double slits is then diffracted, producing a diffraction pattern made up of bright and dark fringes on a screen The typical arrangement of Young's double-slit experiment Diffraction Pattern Constructive interference between light rays forms bright strips, also called fringes , interference fringes or maxima , on the screen Destructive interference forms dark strips, also called dark fringes or minima , on the screen Young's double slit experiment and the resulting diffraction pattern Each bright fringe is identical and has the same width and intensity The fringes are all separated by dark narrow bands of destructive interference The constructive and destructive interference of laser light through a double slit creates bright and dark strips called fringes on a screen placed far away Interference Pattern The Young's double slit interference pattern shows the regions of constructive and destructive interference: Each bright fringe is a peak of equal maximum intensity Each dark fringe is a a trough or minimum of zero intensity The maxima are formed by the constructive interference of light The minima are formed by the destructive interference of light The interference pattern of Young's double-slit diffraction of light When two waves interfere, the resultant wave depends on the path difference between the two waves This extra distance is the path difference The path difference between two waves is determined by the number of wavelengths that cover their difference in length For constructive interference (or maxima), the difference in wavelengths will be an integer number of whole wavelengths For destructive interference (or minima) it will be an integer number of whole wavelengths plus a half wavelength There is usually more than one produced n is the order of the maxima or minima; which represents the position of the maxima away from the central maximum n  = 0 is the central maximum n  = 1 represents the first maximum on either side of the central,  n  = 2 the next one along.... Worked example Determine which orders of maxima are detected at M as the wavelength is increased from 3.5 cm to 12.5 cm. The path difference is more specifically how much longer, or shorter, one path is than the other. In other words, the difference in the distances. Make sure not to confuse this with the distance between the two paths. Fringe Spacing Equation The spacing between the bright or dark fringes in the diffraction pattern formed on the screen can be calculated using the double slit equation: Double slit interference equation with w, s and D represented on a diagram D  is much bigger than any other dimension, normally several metres long s is the separation between the two slits and is often the smallest dimension, normally in mm w  is the distance between the fringes on the screen, often in cm. This can be obtained by measuring the distance between the centre of each consecutive bright spot. The wavelength ,  λ of the incident light increases The distance , s  between the screen and the slits increases The separation , w between the slits decreases Calculate the separation of the two slits. Since w , s and D are all distances, it's easy to mix up which they refer to. Labelling the double-slit diagram with each of these quantities can help ensure you don't use the wrong variable for a quantity. Interference Patterns It is different to that produced by a single slit or a diffraction grating The interference pattern produced when white light is diffracted through a double slit Each maximum is of roughly equal width There are two dark narrow destructive interference fringes on either side All other maxima are composed of a   spectrum The shortest wavelength (violet / blue) would appear   nearest   to the central maximum because it is diffracted the least The longest wavelength (red) would appear   furthest   from the central maximum because it is diffracted the most As the maxima move   further away   from the central maximum, the wavelengths of   blue   observed   decrease   and the wavelengths of   red   observed   increase You've read 0 of your 10 free revision notes Get unlimited access. to absolutely everything: Downloadable PDFs Unlimited Revision Notes Topic Questions Past Papers Model Answers Videos (Maths and Science) Join the 100,000 + Students that ❤️ Save My Exams the (exam) results speak for themselves: Did this page help you? Author: Katie M Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential. 17.1 Understanding Diffraction and Interference Section learning objectives. By the end of this section, you will be able to do the following: Explain wave behavior of light, including diffraction and interference, including the role of constructive and destructive interference in Young’s single-slit and double-slit experiments Perform calculations involving diffraction and interference, in particular the wavelength of light using data from a two-slit interference pattern Teacher Support The learning objectives in this section will help your students master the following standards: (D) investigate behaviors of waves, including reflection, refraction, diffraction, interference, resonance, and the Doppler effect Section Key Terms diffraction Huygens’s principle monochromatic wavefront Diffraction and Interference [BL] Explain constructive and destructive interference graphically on the board. [OL] Ask students to look closely at a shadow. Ask why the edges are not sharp lines. Explain that this is caused by diffraction, one of the wave properties of electromagnetic radiation. Define the nanometer in relation to other metric length measurements. [AL] Ask students which, among speed, frequency, and wavelength, stay the same, and which change, when a ray of light travels from one medium to another. Discuss those quantities in terms of colors (wavelengths) of visible light. We know that visible light is the type of electromagnetic wave to which our eyes responds. As we have seen previously, light obeys the equation where c = 3.00 × 10 8 c = 3.00 × 10 8 m/s is the speed of light in vacuum, f is the frequency of the electromagnetic wave in Hz (or s –1 ), and λ λ is its wavelength in m. The range of visible wavelengths is approximately 380 to 750 nm. As is true for all waves, light travels in straight lines and acts like a ray when it interacts with objects several times as large as its wavelength. However, when it interacts with smaller objects, it displays its wave characteristics prominently. Interference is the identifying behavior of a wave. In Figure 17.2 , both the ray and wave characteristics of light can be seen. The laser beam emitted by the observatory represents ray behavior, as it travels in a straight line. Passing a pure, one-wavelength beam through vertical slits with a width close to the wavelength of the beam reveals the wave character of light. Here we see the beam spreading out horizontally into a pattern of bright and dark regions that are caused by systematic constructive and destructive interference. As it is characteristic of wave behavior, interference is observed for water waves, sound waves, and light waves. That interference is a characteristic of energy propagation by waves is demonstrated more convincingly by water waves. Figure 17.3 shows water waves passing through gaps between some rocks. You can easily see that the gaps are similar in width to the wavelength of the waves and that this causes an interference pattern as the waves pass beyond the gaps. A cross-section across the waves in the foreground would show the crests and troughs characteristic of an interference pattern. Light has wave characteristics in various media as well as in a vacuum. When light goes from a vacuum to some medium, such as water, its speed and wavelength change, but its frequency, f , remains the same. The speed of light in a medium is v = c / n v = c / n , where n is its index of refraction. If you divide both sides of the equation c = f λ c = f λ by n , you get c / n = v = f λ / n c / n = v = f λ / n . Therefore, v = f λ n v = f λ n , where λ n λ n is the wavelength in a medium, and where λ λ is the wavelength in vacuum and n is the medium’s index of refraction. It follows that the wavelength of light is smaller in any medium than it is in vacuum. In water, for example, which has n = 1.333, the range of visible wavelengths is (380 nm)/1.333 to (760 nm)/1.333, or λ n = λ n = 285–570 nm. Although wavelengths change while traveling from one medium to another, colors do not, since colors are associated with frequency. The Dutch scientist Christiaan Huygens (1629–1695) developed a useful technique for determining in detail how and where waves propagate. He used wavefronts , which are the points on a wave’s surface that share the same, constant phase (such as all the points that make up the crest of a water wave). Huygens’s principle states, “Every point on a wavefront is a source of wavelets that spread out in the forward direction at the same speed as the wave itself. The new wavefront is a line tangent to all of the wavelets.” Figure 17.4 shows how Huygens’s principle is applied. A wavefront is the long edge that moves; for example, the crest or the trough. Each point on the wavefront emits a semicircular wave that moves at the propagation speed v . These are drawn later at a time, t , so that they have moved a distance s = v t s = v t . The new wavefront is a line tangent to the wavelets and is where the wave is located at time t . Huygens’s principle works for all types of waves, including water waves, sound waves, and light waves. It will be useful not only in describing how light waves propagate, but also in how they interfere. What happens when a wave passes through an opening, such as light shining through an open door into a dark room? For light, you expect to see a sharp shadow of the doorway on the floor of the room, and you expect no light to bend around corners into other parts of the room. When sound passes through a door, you hear it everywhere in the room and, thus, you understand that sound spreads out when passing through such an opening. What is the difference between the behavior of sound waves and light waves in this case? The answer is that the wavelengths that make up the light are very short, so that the light acts like a ray. Sound has wavelengths on the order of the size of the door, and so it bends around corners. [OL] Discuss the fact that, for a diffraction pattern to be visible, the width of a slit must be roughly the wavelength of the light. Try to give students an idea of the size of visible light wavelengths by noting that a human hair is roughly 100 times wider. If light passes through smaller openings, often called slits, you can use Huygens’s principle to show that light bends as sound does (see Figure 17.5 ). The bending of a wave around the edges of an opening or an obstacle is called diffraction . Diffraction is a wave characteristic that occurs for all types of waves. If diffraction is observed for a phenomenon, it is evidence that the phenomenon is produced by waves. Thus, the horizontal diffraction of the laser beam after it passes through slits in Figure 17.2 is evidence that light has the properties of a wave. Once again, water waves present a familiar example of a wave phenomenon that is easy to observe and understand, as shown in Figure 17.6 . Watch Physics Single-slit interference. This video works through the math needed to predict diffraction patterns that are caused by single-slit interference. Which values of m denote the location of destructive interference in a single-slit diffraction pattern? whole integers, excluding zero whole integers real numbers excluding zero real numbers The fact that Huygens’s principle worked was not considered enough evidence to prove that light is a wave. People were also reluctant to accept light’s wave nature because it contradicted the ideas of Isaac Newton, who was still held in high esteem. The acceptance of the wave character of light came after 1801, when the English physicist and physician Thomas Young (1773–1829) did his now-classic double-slit experiment (see Figure 17.7 ). When light passes through narrow slits, it is diffracted into semicircular waves, as shown in Figure 17.8 (a). Pure constructive interference occurs where the waves line up crest to crest or trough to trough. Pure destructive interference occurs where they line up crest to trough. The light must fall on a screen and be scattered into our eyes for the pattern to be visible. An analogous pattern for water waves is shown in Figure 17.8 (b). Note that regions of constructive and destructive interference move out from the slits at well-defined angles to the original beam. Those angles depend on wavelength and the distance between the slits, as you will see below. Virtual Physics Wave interference. This simulation demonstrates most of the wave phenomena discussed in this section. First, observe interference between two sources of electromagnetic radiation without adding slits. See how water waves, sound, and light all show interference patterns. Stay with light waves and use only one source. Create diffraction patterns with one slit and then with two. You may have to adjust slit width to see the pattern. Visually compare the slit width to the wavelength. When do you get the best-defined diffraction pattern? when the slit width is larger than the wavelength when the slit width is smaller than the wavelength when the slit width is comparable to the wavelength when the slit width is infinite Calculations Involving Diffraction and Interference [BL] The Greek letter θ θ is spelled theta . The Greek letter λ λ is spelled lamda . Both are pronounced the way you would expect from the spelling. The plurals of maximum and minimum are maxima and minima , respectively. [OL] Explain that monochromatic means one color. Monochromatic also means one frequency . The sine of an angle is the opposite side of a right triangle divided by the hypotenuse. Opposite means opposite the given acute angle. Note that the sign of an angle is always ≥ 1. The fact that the wavelength of light of one color, or monochromatic light, can be calculated from its two-slit diffraction pattern in Young’s experiments supports the conclusion that light has wave properties. To understand the basis of such calculations, consider how two waves travel from the slits to the screen. Each slit is a different distance from a given point on the screen. Thus different numbers of wavelengths fit into each path. Waves start out from the slits in phase (crest to crest), but they will end up out of phase (crest to trough) at the screen if the paths differ in length by half a wavelength, interfering destructively. If the paths differ by a whole wavelength, then the waves arrive in phase (crest to crest) at the screen, interfering constructively. More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths ( 1 2 λ ,   3 2 λ ,   5 2 λ ,  etc .) ( 1 2 λ ,   3 2 λ ,   5 2 λ ,  etc .) , then destructive interference occurs. Similarly, if the paths taken by the two waves differ by any integral number of wavelengths ( λ ,   2 λ ,   3 λ ,  etc .) ( λ ,   2 λ ,   3 λ ,  etc .) , then constructive interference occurs. Figure 17.9 shows how to determine the path-length difference for waves traveling from two slits to a common point on a screen. If the screen is a large distance away compared with the distance between the slits, then the angle θ θ between the path and a line from the slits perpendicular to the screen (see the figure) is nearly the same for each path. That approximation and simple trigonometry show the length difference, Δ L Δ L , to be d sin θ d sin θ , where d is the distance between the slits, To obtain constructive interference for a double slit, the path-length difference must be an integral multiple of the wavelength, or Similarly, to obtain destructive interference for a double slit, the path-length difference must be a half-integral multiple of the wavelength, or The number m is the order of the interference. For example, m = 4 is fourth-order interference. Figure 17.10 shows how the intensity of the bands of constructive interference decreases with increasing angle. Light passing through a single slit forms a diffraction pattern somewhat different from that formed by double slits. Figure 17.11 shows a single-slit diffraction pattern. Note that the central maximum is larger than those on either side, and that the intensity decreases rapidly on either side. The analysis of single-slit diffraction is illustrated in Figure 17.12 . Assuming the screen is very far away compared with the size of the slit, rays heading toward a common destination are nearly parallel. That approximation allows a series of trigonometric operations that result in the equations for the minima produced by destructive interference. When rays travel straight ahead, they remain in phase and a central maximum is obtained. However, when rays travel at an angle θ θ relative to the original direction of the beam, each ray travels a different distance to the screen, and they can arrive in or out of phase. Thus, a ray from the center travels a distance λ / 2 λ / 2 farther than the ray from the top edge of the slit, they arrive out of phase, and they interfere destructively. Similarly, for every ray between the top and the center of the slit, there is a ray between the center and the bottom of the slit that travels a distance λ / 2 λ / 2 farther to the common point on the screen, and so interferes destructively. Symmetrically, there will be another minimum at the same angle below the direct ray. Below we summarize the equations needed for the calculations to follow. The speed of light in a vacuum, c , the wavelength of the light, λ λ , and its frequency, f , are related as follows. The wavelength of light in a medium, λ n λ n , compared to its wavelength in a vacuum, λ λ , is given by To calculate the positions of constructive interference for a double slit, the path-length difference must be an integral multiple, m , of the wavelength. λ λ where d is the distance between the slits and θ θ is the angle between a line from the slits to the maximum and a line perpendicular to the barrier in which the slits are located. To calculate the positions of destructive interference for a double slit, the path-length difference must be a half-integral multiple of the wavelength: For a single-slit diffraction pattern, the width of the slit, D , the distance of the first ( m = 1) destructive interference minimum, y , the distance from the slit to the screen, L , and the wavelength, λ λ , are given by Also, for single-slit diffraction, where θ θ is the angle between a line from the slit to the minimum and a line perpendicular to the screen, and m is the order of the minimum. Worked Example Two-slit interference. Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm, and you find that the third bright line on a screen is formed at an angle of 10.95º relative to the incident beam. What is the wavelength of the light? The third bright line is due to third-order constructive interference, which means that m = 3. You are given d = 0.0100 mm and θ θ = 10.95º. The wavelength can thus be found using the equation d sin θ = m λ d sin θ = m λ for constructive interference. The equation is d sin θ = m λ d sin θ = m λ . Solving for the wavelength, λ λ , gives Substituting known values yields To three digits, 633 nm is the wavelength of light emitted by the common He-Ne laser. Not by coincidence, this red color is similar to that emitted by neon lights. More important, however, is the fact that interference patterns can be used to measure wavelength. Young did that for visible wavelengths. His analytical technique is still widely used to measure electromagnetic spectra. For a given order, the angle for constructive interference increases with λ λ , so spectra (measurements of intensity versus wavelength) can be obtained. Single-Slit Diffraction Visible light of wavelength 550 nm falls on a single slit and produces its second diffraction minimum at an angle of 45.0° relative to the incident direction of the light. What is the width of the slit? From the given information, and assuming the screen is far away from the slit, you can use the equation D sin θ = m λ D sin θ = m λ to find D . Quantities given are λ λ = 550 nm, m = 2, and θ 2 θ 2 = 45.0°. Solving the equation D sin θ = m λ D sin θ = m λ for D and substituting known values gives You see that the slit is narrow (it is only a few times greater than the wavelength of light). That is consistent with the fact that light must interact with an object comparable in size to its wavelength in order to exhibit significant wave effects, such as this single-slit diffraction pattern. Practice Problems What is the width of a single slit through which 610-nm orange light passes to form a first diffraction minimum at an angle of 30.0°? Check Your Understanding Use these problems to assess student achievement of the section’s learning objectives. If students are struggling with a specific objective, these problems will help identify which and direct students to the relevant topics. The wavelength first decreases and then increases. The wavelength first increases and then decreases. The wavelength increases. The wavelength decreases. This is a diffraction effect. Your whole body acts as the origin for a new wavefront. This is a diffraction effect. Every point on the edge of your shadow acts as the origin for a new wavefront. This is a refraction effect. Your whole body acts as the origin for a new wavefront. This is a refraction effect. Every point on the edge of your shadow acts as the origin for a new wavefront. This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission. Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute Texas Education Agency (TEA). The original material is available at: https://www.texasgateway.org/book/tea-physics . Changes were made to the original material, including updates to art, structure, and other content updates. Access for free at https://openstax.org/books/physics/pages/1-introduction Authors: Paul Peter Urone, Roger Hinrichs Publisher/website: OpenStax Book title: Physics Publication date: Mar 26, 2020 Location: Houston, Texas Book URL: https://openstax.org/books/physics/pages/1-introduction Section URL: https://openstax.org/books/physics/pages/17-1-understanding-diffraction-and-interference © Jun 7, 2024 Texas Education Agency (TEA). The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University. IIT JEE Study Material Youngs Double Slit Experiment Young's Double Slit Experiment What is young’s double slit experiment. Young’s double slit experiment uses two coherent sources of light placed at a small distance apart. Usually, only a few orders of magnitude greater than the wavelength of light are used. Young’s double slit experiment helped in understanding the wave theory of light , which is explained with the help of a diagram. As shown, a screen or photodetector is placed at a large distance, ‘D’, away from the slits. Download Complete Chapter Notes of Wave Optics Download Now JEE Main 2021 LIVE Physics Paper Solutions 24 Feb Shift-1 Memory-based The original Young’s double slit experiment used diffracted light from a single source passed into two more slits to be used as coherent sources. Lasers are commonly used as coherent sources in modern-day experiments. Table of Contents Position of Fringes Shape of Fringes Intensity of Fringes Special Cases Displacement of Fringes Each source can be considered a source of coherent light waves . At any point on the screen at a distance ‘y’ from the centre, the waves travel distances  l 1  and  l 2 to create a path difference of Δl at the point. The point approximately subtends an angle of θ at the sources (since the distance D is large, there is only a very small difference between the angles subtended at sources). Derivation of Young’s Double Slit Experiment Consider a monochromatic light source ‘S’ kept at a considerable distance from two slits: s 1 and s 2 . S is equidistant from s 1 and s 2 . s 1 and s 2 behave as two coherent sources as both are derived from S. The light passes through these slits and falls on a screen which is at a distance ‘D’ from the position of slits s 1 and s 2 . ‘d’ is the separation between two slits. If s 1 is open and s 2 is closed, the screen opposite to s 1 is closed, and only the screen opposite to s 2 is illuminated. The interference patterns appear only when both slits s 1 and s 2 are open. When the slit separation (d) and the screen distance (D) are kept unchanged, to reach P, the light waves from s 1 and s 2 must travel different distances. It implies that there is a path difference in Young’s double slit experiment between the two light waves from s 1 and s 2 . Approximations in Young’s double slit experiment  Approximation  1: D > > d: Since D > > d, the two light rays are assumed to be parallel. Approximation  2:   d/λ >> 1:  Often, d is a fraction of a millimetre, and λ is a fraction of a micrometre for visible light. Under these conditions, θ is small. Thus, we can use the approximation sin θ = tan θ ≈ θ = λ/d. ∴ path difference, Δz = λ/d This is the path difference between two waves meeting at a point on the screen. Due to this path difference in Young’s double slit experiment, some points on the screen are bright, and some points are dark. Now, we will discuss the position of these light and dark fringes and fringe width. Position of Fringes in Young’s Double Slit Experiment Position of bright fringes. For maximum intensity or bright fringe to be formed at P, Path difference, Δz = nλ (n = 0, ±1, ±2, . . . .) i.e., xd/D = nλ The distance of the n th   bright fringe from the centre is x n = nλD/d Similarly, the distance of the (n-1) th  bright fringe from the centre is x (n-1) = (n -1)λD/d Fringe width,  β =  x n   – x (n-1)  = nλD/d – (n -1)λD/d = λD/d (n = 0, ±1, ±2, . . . .) Position of Dark Fringes For minimum intensity or dark fringe to be formed at P, Path difference, Δz = (2n + 1) (λ/2) (n = 0, ±1, ±2, . . . .) i.e., x = (2n +1)λD/2d The distance of the n th   dark fringe from the centre is x n = (2n+1)λD/2d x (n-1) = (2(n-1) +1)λD/2d Fringe width, β = x n   – x (n-1)  = (2n + 1) λD/2d – (2(n -1) + 1)λD/2d = λD/d Fringe Width The distance between two adjacent bright (or dark) fringes is called the fringe width. If the apparatus of Young’s double slit experiment is immersed in a liquid of refractive index  (μ), then the wavelength of light and fringe width decreases ‘μ’ times. If white light is used in place of monochromatic light, then coloured fringes are obtained on the screen, with red fringes larger in size than violet. Angular Width of Fringes Let the angular position of n th bright fringe is θ n, and because of its small value, tan θ n   ≈ θ n Similarly, the angular position of (n+1) th bright fringe is θ n+1,  then ∴ The angular width of a fringe in Young’s double slit experiment is given by, Angular width is independent of ‘n’, i.e., the angular width of all fringes is the same. Maximum Order of Interference Fringes But ‘n’ values cannot take infinitely large values as it would violate the 2 nd approximation. i.e., θ is small (or) y < < D When the ‘n’ value becomes comparable to d/ λ, path difference can no longer be given by d γ/D. Hence for maxima, path difference = nλ The above represents the box function or greatest integer function. Similarly, the highest order of interference minima The Shape of Interference Fringes in YDSE From the given YDSE diagram, the path difference between the two slits is given by The above equation represents a hyperbola with its two foci as, s 1 and s 2 . The interference pattern we get on the screen is a section of a hyperbola when we revolve the hyperbola about the axis s 1 s 2 . If the screen is a yz plane, fringes are hyperbolic with a straight central section. If the screen is xy plane , the fringes are hyperbolic with a straight central section. The Intensity of Fringes in Young’s Double Slit Experiment For two coherent sources, s 1 and s 2 , the resultant intensity at point p is given by I = I 1 + I 2 + 2 √(I 1 . I 2 ) cos φ Putting I 1 = I 2 = I 0  (Since, d<<<D) I = I 0 + I 0 + 2 √(I 0 .I 0 ) cos φ I = 2I 0 + 2 (I 0 ) cos φ I = 2I 0 (1 +  cos φ) For maximum intensity phase difference   φ = 2nπ Then, path difference \(\begin{array}{l}\Delta x=\frac{\lambda }{{2}{\pi }}\left( {2}n{\pi } \right)\end{array} \) = nλ The intensity of bright points is maximum and given by I max = 4I 0 For minimum intensity φ = (2n – 1) π Phase difference φ = (2n – 1)π Thus, the intensity of minima is given by If I 1 ≠ I 2 , I min ≠ 0. Rays Not Parallel to Principal Axis: From the above diagram, Using this, we can calculate different positions of maxima and minima. Source Placed beyond the Central Line: If the source is placed a little above or below this centre line, the wave interaction with S 1 and S 2 has a path difference at point P on the screen. Δ x= (distance of ray 2) – (distance of ray 1) = bd/a + yd/D → (*) We know Δx = nλ for maximum Δx = (2n – 1) λ/2 for minimum By knowing the value of Δx from (*), we can calculate different positions of maxima and minima . Displacement of Fringes in YDSE When a thin transparent plate of thickness ‘t’ is introduced in front of one of the slits in Young’s double slit experiment, the fringe pattern shifts toward the side where the plate is present. The dotted lines denote the path of the light before introducing the transparent plate. The solid lines denote the path of the light after introducing a transparent plate. Where μt is the optical path. Then, we get, Term (1) defines the position of a bright or dark fringe; term (2) defines the shift that occurred in the particular fringe due to the introduction of a transparent plate. Constructive and Destructive Interference For constructive interference, the path difference must be an integral multiple of the wavelength. Thus, for a bright fringe to be at ‘y’, Or, y = nλD/d Where n = ±0,1,2,3….. The 0th fringe represents the central bright fringe. Similarly, the expression for a dark fringe in Young’s double slit experiment can be found by setting the path difference as Δl = (2n+1)λ/2 This simplifies to (2n+1)λ/2 = y d/D y = (2n+1)λD/2d Young’s double slit experiment was a watershed moment in scientific history because it firmly established that light behaved like a wave. The double slit experiment was later conducted using electrons , and to everyone’s surprise, the pattern generated was similar as expected with light. This would forever change our understanding of matter and particles, forcing us to accept that matter, like light, also behaves like a wave. Wave Optics Young’s double slit experiment. Frequently Asked Questions on Young’s Double Slit Experiment What was the concept explained by young’s double slit experiment. Young’s double slit experiment helps in understanding the wave theory of light. What are the formulas derived from Young’s double slit experiment? For constructive interference, dsinθ = mλ , for m = 0,1,-1,2,-2 For destructive interference, dsinθ = (m+½)λ, for m = 0,1,-1,2,-2 Here, d is the distance between the slits. λ is the wavelength of the light waves. What is called a fringe width? The distance between consecutive bright or dark fringe is called the fringe width. What kind of source is used in Young’s double slit experiment? A coherent source is used in Young’s double slit experiment. Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin! Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz Visit BYJU’S for all JEE related queries and study materials Your result is as below Request OTP on Voice Call Leave a Comment Cancel reply Your Mobile number and Email id will not be published. Required fields are marked * Post My Comment Register with Aakash BYJU'S & Download Free PDFs Register with byju's & watch live videos. 294 IMAGES Young's double slit experiment diagram. Interference of light waves Interference of Light Waves & Young's Experiment: Videos and Examples Huygens Principle Interference of Light Waves and Young's Experiment Interference and diffraction of light waves. Young’s experiment. Newton Deriving Young’s Double Slit Interference Formulas Interference: Young’s Double Slit Experiment of Optics in Physics Class 12 VIDEO Interference Of Light Wave Optics || Lec 05 || Physics Class 12 || i-SCIENCE Tamil interference of light Waves and Young's experiment Interference of light waves and Young's double slit experiment// PART 1/Ch10 Wave Optics //Class 12 Young's double slit diffraction [IB Physics SL/HL] Young's Double Slit Experiment derivation| Wave Optics class 12| boards|JEE|maxima and minima points COMMENTS Young's interference experiment Thomas Young's sketch of interference based on observations of water waves [6] In 1801, Young presented a famous paper to the Royal Society entitled "On the Theory of Light and Colours" [7] which describes various interference phenomena. In 1803, he described his famous interference experiment. [8] Young's Double Slits Experiment Derivation Introduction To Young's Double Slits Experiment. During the year 1801, Thomas Young carried out an experiment where the wave and particle nature of light and matter were demonstrated. The schematic diagram of the experimental setup is shown below-Figure(1): Young double slit experimental set up along with the fringe pattern. PDF Chapter 14 Interference and Diffraction 14.2 Young's Double-Slit Experiment In 1801 Thomas Young carried out an experiment in which the wave nature of light was demonstrated. The schematic diagram of the double-slit experiment is shown in Figure 14.2.1. Figure 14.2.1 Young's double-slit experiment. A monochromatic light source is incident on the first screen which contains a slit . Young's Double Slit Experiment Section Summary. Young's double slit experiment gave definitive proof of the wave character of light. An interference pattern is obtained by the superposition of light from two slits. There is constructive interference when d sin θ = mλ ( for m = 0, 1, −1, 2, −2, . . . ), where d is the distance between the slits, θ is the angle ... Young's Double Slit Experiment The story of Young's Double Slit Experiment begins in the early 19th century with a physicist named Thomas Young. At that time, the nature of light was a ... Interference of Light Waves. ... This setup and the resulting pattern provide a clear demonstration of the fundamental principle of interference in wave optics. Derivation for YDSE. 27.3 Young's Double Slit Experiment The acceptance of the wave character of light came many years later when, in 1801, the English physicist and physician Thomas Young (1773-1829) did his now-classic double slit experiment (see Figure 27.10). Young's experiment interference. Young's experiment, classical investigation into the nature of light, an investigation that provided the basic element in the development of the wave theory and was first performed by the English physicist and physician Thomas Young in 1801. In this experiment, Young identified the phenomenon called interference. Light Light - Wave, Interference, Diffraction: The observation of interference effects definitively indicates the presence of overlapping waves. Thomas Young postulated that light is a wave and is subject to the superposition principle; his great experimental achievement was to demonstrate the constructive and destructive interference of light (c. 1801). Interference of Light Waves and Young's Experiment Condition of a Steady Interference Pattern. A 1 = A 2 . The amplitude of two waves must be equal. λ 1 = λ 2. The two waves interfering must have same color i.e they must be of the same wavelength. Sources must be narrow. The distance between source should be less. Source and screen should be at large distance. We should get coherent sources. 27.3 Young's Double Slit Experiment Section Summary. Young's double slit experiment gave definitive proof of the wave character of light. An interference pattern is obtained by the superposition of light from two slits. There is constructive interference when d sin θ = mλ (for m = 0, 1, −1, 2, −2,…) d sin θ = m λ ( for m = 0, 1, − 1, 2, − 2, …), where d d is the ... Young's double slit experiment derivation Young's double slit experiment derivation. December 18, 2022 by shabbusharma. One of the first demonstration of the intererference of light waves was given by Young - an English physicist in 1801. We have learnt that two essential conditions to obtain an interference phenomenon are : Interference of light Young's Double Slit Experiment. The great scientist Young performed an experiment to prove the wave nature of light by explaining the phenomenon of interference of light. In Young's double slit experiment, two coherent sources were generated using diffracted light from a single slit. Note that the waves must have a constant phase difference ... Young's Double Slit Interference Young's Double Slit Interference. Waves can be added together either constructively or destructively. The result of adding two waves of the same frequency depends on the value of the phase of the wave at the point in which the waves are added. Electromagnetic waves are subject to interference. Young's double slit equation Let's derive a formula that relates all the variables in Young's double slit experiment. Created by David SantoPietro.Watch the next lesson: https://www.khan... Young's Double-Slit Experiment Double Slit Interference. Young's double-slit experiment produces a diffraction and an interference pattern using either: The interference of two coherent wave sources. A single wave source passing through a double slit. In this typical set-up for Young's double slit experiment: The laser light source is placed behind the single slit. Khan Academy What is the phenomenon of interference of light waves and how can we observe it using a simple experiment? Watch this video to learn about Young's double slit introduction, a classic demonstration of the wave nature of light. You will also find out how to calculate the distance between the bright fringes on a screen using Young's double slit equation. If you want to learn more, you can check ... Double-slit experiment In modern physics, the double-slit experiment demonstrates that light and matter can satisfy the seemingly incongruous classical definitions for both waves and particles. This ambiguity is considered evidence for the fundamentally probabilistic nature of quantum mechanics.This type of experiment was first performed by Thomas Young in 1801, as a demonstration of the wave behavior of visible ... Khan Academy Khanmigo is now free for all US educators! Plan lessons, develop exit tickets, and so much more with our AI teaching assistant. 4.3 Double-Slit Diffraction In other words, the locations of the interference fringes are given by the equation d sin θ = m λ d sin θ = m λ, the same as when we considered the slits to be point sources, but the intensities of the fringes are now reduced by diffraction effects, according to Equation 4.4. [Note that in the chapter on interference, we wrote d sin θ = m λ d sin θ = m λ and used the integer m to refer ... 17.1 Understanding Diffraction and Interference where c = 3.00 × 10 8 c = 3.00 × 10 8 m/s is the speed of light in vacuum, f is the frequency of the electromagnetic wave in Hz (or s -1), and λ λ is its wavelength in m. The range of visible wavelengths is approximately 380 to 750 nm. As is true for all waves, light travels in straight lines and acts like a ray when it interacts with objects several times as large as its wavelength. Young's Double Slit Experiment When the slit separation (d) and the screen distance (D) are kept unchanged, to reach P, the light waves from s 1 and s 2 must travel different distances. It implies that there is a path difference in Young's double slit experiment between the two light waves from s 1 and s 2. Approximations in Young's double slit experiment essay homework paper phd review speech study thesis