ydse experiment formula

• Derivation of Physics Formula
• Youngs Double Slit Experiment Derivation

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-

A beam of monochromatic light is made incident on the first screen, which contains the slit S 0 . The emerging light then incident on the second screen which consists of two slits namely, S 1 , S 2 . These two slits serve as a source of coherent light. The emerging light waves from these slits interfere to produce an interference pattern on the screen. The interference pattern consists of consecutive bright and dark fringes. The dark fringes are the result of destructive interference and bright fringes are the result of constructive interference.

You may also want to check out these topics given below!

• Superposition of waves
• Diffraction
• Polarisation by scattering

• The light falls on the screen at the point P. which is at a distance y from the centre O.
• The distance between the double-slit system and the screen is L
• The two slits are separated by the distance d
• Distance travelled by the light ray from slit 1 to point P on the screen is r 1
• Distance travelled by the light ray from slit 2 to point P on the screen is r 2
• Thus, the light ray from slit 2 travels an extra distance of ẟ = r 2 -r 1 than light ray from slit 1.
• This extra distance is termed as Path difference.

Refer to Figure(3) Applying laws of cosines; we can write –

∵ Cos(90° – θ) = Sinθ

∵ Cos(90° + θ) = -Sinθ

Subtracting equation (2) from (1) we get-

Let us impose the limit that the distance between the double slit system and the screen is much greater than the distance between the slits. That is L >> d. The sum of r 1 and r 2 can be approximated to r 1 + r 2 ≅2r. Thus, the path difference becomes –

In this limit, the two rays r 1 and r 2 are essentially treated as parallel. (See Figure(4))

Figure(4): Assuming L >> d, The path difference between two rays.

To compare the phase of two waves, the value of path difference (ẟ) plays a crucial role.

Constructive interference is seen when path difference (𝛿) is zero or integral multiple of the wavelength (λ).

Where m is order number. The Zeroth order maximum (m=0)corresponds to the central bright fringe, here 𝜃=0. The first order maxima(m=±⁤1)(bright fringe) are on either side the central fringe.

Similarly, when 𝛿 is an odd integral multiple of λ/2, the resultant fringes will be 180 0 out of phase, thus, forming a dark fringe.

Assuming the distance between the slits are much greater than the wavelength of the incident light, we get-

Substituting it in the constructive and destructive interference condition we can get the position of bright and dark fringes, respectively. The equation is as follows-

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27.3 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 27.10 ).

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 λ ) light to clarify the effect. Figure 27.11 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 27.12 (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 27.12 (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 27.13 . 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 27.13 (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 27.13 (b). More generally, if the paths taken by the two waves differ by any half-integral number of wavelengths [ ( 1 / 2 ) λ ( 1 / 2 ) λ , ( 3 / 2 ) λ ( 3 / 2 ) λ , ( 5 / 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 λ , 2 λ 2 λ , 3 λ 3 λ , 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 27.14 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 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 d sin θ , where d 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

where λ λ is the wavelength of the light, d d is the distance between slits, and θ θ is the angle from the original direction of the beam as discussed above. We call m m the order of the interference. For example, m = 4 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 27.15 . 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 λ λ and m m , the smaller d d is, the larger θ θ must be, since sin θ = mλ / d sin θ = mλ / d . This is consistent with our contention that wave effects are most noticeable when the object the wave encounters (here, slits a distance d d apart) is small. Small d d gives large θ θ , hence a large effect.

Example 27.1

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 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 m = 3 . We are given d = 0 . 0100 mm d = 0 . 0100 mm and θ = 10 . 95º θ = 10 . 95º . The wavelength can thus be found using the equation d sin θ = mλ d sin θ = mλ for constructive interference.

The equation is d sin θ = mλ d sin θ = mλ . Solving for the wavelength λ λ 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 λ λ , so that spectra (measurements of intensity versus wavelength) can be obtained.

Example 27.2

Calculating highest order possible.

Interference patterns do not have an infinite number of lines, since there is a limit to how big m 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, … ) d sin θ = mλ (for m = 0, 1, − 1, 2, − 2, … ) describes constructive interference. For fixed values of d d and λ λ , the larger m m is, the larger sin θ sin θ is. However, the maximum value that sin θ sin θ can have is 1, for an angle of 90º 90º . (Larger angles imply that light goes backward and does not reach the screen at all.) Let us find which m m corresponds to this maximum diffraction angle.

Solving the equation d sin θ = mλ d sin θ = mλ  for  m m gives

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

Therefore, the largest integer m m 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.

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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

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

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Young's Double Slit Experiment

Interference occurs when two coherent waves of same frequency but different phase superpose. \begin{align} &E_1=E_{01}\sin(kx-\omega t)\\ &E_2=E_{02}\sin(kx-\omega t+\delta)\nonumber\\ &E=E_1+E_2=E_0\sin(kx-\omega t+\epsilon) \nonumber \\ & E_0=\sqrt{{E_{01}}^2 + {E_{02}}^2 + 2 E_{01} E_{02}\cos\delta} \nonumber\\ &\tan\epsilon=\frac{E_{02\sin\delta}}{E_{01}+E_{02}\cos\delta}\nonumber \end{align}

The path difference between interferening waves in a Young's double slit experiment is given by \begin{align} \Delta x=\frac{dy}{D} \end{align}

The phase difference is given by \begin{align} \delta=\frac{2\pi}{\lambda}\Delta x \end{align}

The conditions for constructive and destructive interference are: for integer $n$, \begin{align} \delta=\left\{\begin{array}{ll} 2n\pi, & \text{constructive};\\ (2n+1)\pi, & \text{destructive}, \end{array}\right.\nonumber \end{align} \begin{align} \Delta x=\left\{\begin{array}{ll} n\lambda, & \text{constructive}; \\ \left(n+\tfrac{1}{2}\right)\lambda, & \text{destructive} \end{array}\right.\nonumber \end{align}

The fringe width is given by \begin{align} w=\frac{\lambda D}{d} \end{align}

• In the Young’s double slit experiment the size of the apertures forming the two coherent sources is given as 0.02 mm. Plot the intensity pattern seen on the screen if the distance between the apertures is kept (a) 0.09 mm (b) 0.03 mm.

Problems from IIT JEE

• $\beta_G > \beta_B > \beta_R$
• $\beta_B > \beta_G > \beta_R$
• $\beta_R > \beta_B > \beta_G$
• $\beta_R > \beta_G > \beta_B$

Solution: The fringe width $\beta$ is related to the wavelength $\lambda$ by, $\beta={\lambda D}/{d}$. In the visible spectrum $\lambda_B \beta_G>\beta_B$.

• the intensities of both maxima and the minima increase.
• the intensity of the maxima increases and the minima has zero intensity.
• the intensity of maxima decreases and that of minima increases.
• the intensity of maxima decreases and the minima has zero intensity.

Solution: Let the intensity of light passing through the two slits be $I_1$ and $I_2$. The intensity at a point having a phase difference $\delta$ is given by, \begin{align} I=I_1+I_2+2\sqrt{I_1I_2}\cos\delta. \end{align} The above equation gives the intensity at maxima ($\delta=0$) and minima ($\delta=\pi$) as, \begin{align} &I_\text{max}=I_1+I_2+2\sqrt{2I_1I_2},\\ &I_\text{min}=I_1+I_2-2\sqrt{I_1I_2}. \end{align} When the slits are of same width, $I_1=I_0$, $I_2=I_0$, $I_\text{max}=4I_0$ and $I_\text{min}=0$. When the width of one of the slit is doubled, \begin{align} I_1&=I_0,\\ I_2&=2I_0,\\ I_\text{max}&=(3+2\sqrt{2})I_0=5.83I_0\\ I_\text{min}&=(3-2\sqrt{2})I_0=0.17I_0. \end{align} The readers are encouraged to show that both $I_\text{max}$ and $I_\text{min}$ increase if $I_2=xI_0$ for some $x>1$.

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Young’s Double Slit Experiment

Conditions for interference.

(a) The two sources should emit, continuously, waves of some wave-length or frequency. While deriving conditions for maxima and minima, we have taken ‘I’ for both the waves to be same. 

(b) The amplitudes of the two waves should be either or nearly equal. A good contrast between a maxima and minima can only be obtained if the amplitudes of two waves are equal or nearly equal.

(c) The two sources should be narrow. A broader source can be supposed to be a combination of a number of narrow sources assembled side-by-side. Interference patterns due to these narrow sources may overlap each other.

(d) The sources should be close to each other. The fringe width varies inversely as distance ‘d’ between the two sources. So, interference pattern will be more clear and distant if ‘d’ is small.

(e) The two sources should be coherent one.  

Double Slit Experiment

The phenomenon of interference was first observed and demonstrated by Thomas Young in 1801. The experimental set up is shown in figure.

Light from a narrow slit S, illuminated by a monochromatic source, is allowed to fall on two narrow slits A and B placed very close to each other. The width of each slit is about 0.03 mm and they are about 0.3 mm apart. Since A and B are equidistant from S, light waves from S reach A and B in phase. So A and B acts as coherent sources.

When one of the slits is covered, the fringes disappear and there is uniform illumination on the screen. This shows clearly that the bands are due to interference.

Let d be the distance between two coherent sources A and B of wavelength λ. A screen XY is placed parallel to AB at a distance D from the coherent sources. C is the midpoint of AB. O is a point on the screen equidistant from A and B. P is a point at a distance x from O, as shown in Fig 5.17. Waves from A and B meet at P in phase or out of phase depending upon the path difference between two waves

Draw AM perpendicular to BP

The path difference δ = BP – AP

δ = BP – AP = BP – MP = BM

The path difference δ = θ.d 

In right angled triangle COP, tan θ = OP/CO = x/D

For small values of θ, tan θ = θ

Thus, the path difference δ = xd/D  

Bright Fringes

By the principle of interference, condition for constructive interference is the path difference = nλ

Here, n = 0,1,2.....indicate the order of bright fringes

So, x = (D/d) nλ

This equation gives the distance of the n th bright fringe from the point O.

Dark Fringes

Here, n = 1,2,3 … indicate the order of the dark fringes.

So, x =  (D/d) [(2n – 1)λ/2]

This equation gives the distance of the n th dark fringe from the point O. Thus, on the screen alternate dark and bright bands are seen on either side of the central bright band.

Band Width (β)

The distance between any two consecutive bright or dark bands is called bandwidth.

The distance between (n+1) th and n th order consecutive bright fringes from O is given by,

x n+1  – x n  =  [(D/d) [(n+1)λ] –  (D/d) [(n)λ]] = (D/d) λ

Bandwidth,   β = (D/d) λ

Similarly, it can be proved that the distance between two consecutive dark bands is also equal to (D/d) λ. Since bright and dark fringes are of same width, they are equi−spaced on either side of central maximum.  

Condition for Obtaining Clear and Broad Interference Bands

The screen should be as far away from the source as possible.

The wavelength of light used must be larger.

The two coherent sources must be as close as possible.  

Watch this Video for more reference  

The interference pattern in which the positions of maximum and minimum intensity of light remain fixed with time, is called sustained or permanent interference pattern. The conditions for the formation of sustained interference may be stated as :

(a) The two sources should be coherent

(b)  Two sources should be very narrow

(c)  The sources should lie very close to each other to form distinct and broad fringes.

Problem (JEE Advanced):

In Young’s double slit experiment, the intensities at two points P 1  and P 2  on the screen are respectively I 1  and I 2 . If P 1  is located at the centre of a bright fringe and P 2  is located at a distance equal to a quarter of fringe width from P 1 , then find I 1 /I 2 .

Here, y = ω = λD/4d

?x = yd/D = λ/4

φ = (2π/λ)?x = π/2

φ/2 =  π/4

Now, I 2  = I 1  cos 2 (φ/2)

Or, I 1 / I 2  = 1/cos 2 (φ/2) = 2

From the above observation we conclude that, the ratio of  I 1 /I 2  would be 2.

A beam of light consisting of two wavelengths 6500  o A and 5200  o A is used to obtain interference fringes. The distance between the slits is 2.0 mm and the distance between the plane of the slits and the screen is 120 cm.

(a)    Find the distance of the third bright fringe on the screen from the central maxima for the wavelength 6500  o A.

(b)    What is the least distance from the central maxima where the bright fringes due to both the wavelengths coincide?

(i) y 3  = n. Dλ/d = 3 x 1.2m x 6500 x 10 -10 m / 2 x 10 -3 m  = 0.12cm

Let nth maxima of light with wavelength 6500 Å coincides with that of mth maxima of 5200Å.

(ii) m x 6500A o  x D/d = n x 5200A o  x D/d ⇒ m/n = 5200/6500 = 4/5

Least distance = y 4  = 4.D (6500A o )/d = 4 x 6500 x 10 -10  x 1.2/ 2 x 10 -3 m = 0.16cm

In Young’s experiment be performed with two slits in water instead of in air:

(a) The fringes will be smaller in number          (b) The fringes will be broader

(c) The fringes will be narrower                        (d) No fringes will be obtained

In Young’s interference experiment with one source and two slits, one slit is covered with a cello phone sheet so that half the intensity is absorbed. Then:

(a) No fringe is obtained

(b) Bright fringe will be brighter and dark fringe is darker

(c) All fringes will be darker

(d) Bright fringes will be less bright and dark fringes will be less dark.

Question 3   [IIT 1981]

In Young’s interference experiment, the separation between the slits is halved and the distance between slits and the screen is doubled. The fringe width is: 

(a) Unchanged       (b) Halved

(c) Doubled           (d) Quadrupled

In   Young’s experiment for interference of light with two slits, reinforcement takes place when sin θ = mλ/d, d is,

(a) distance from slit to screen          (b) distance between dark and bright fringe

(c) distance between the slits           (d) width of m th fringe.

In Young’s double slit experiment, the distance between the slits is gradually increased. The width of the fringes system:

(a) increases           (b) decreases

(c) remains the same   (d) halved

Related Resources

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Intensity in Young’s Double Slit Experiment - Important Topic for JEE

• Intensity In Youngs Double Slit Experiment

An Introduction to Young’s Double Slit Experiment

The Young’s Double Slit Experiment, or YDSE, is one of the most important experiments that laid the foundations for Modern Physics. It is an experiment that demonstrates the interference effect of light from two slits. This experiment showed the wave nature of light by showing that light interferes with itself and produces interference patterns. There has been a lot of debate about the nature of light, and Newton himself believed that light had a corpuscular nature. Huygens thought that light was a wave. However, due to Newton’s stature at that time, his view prevailed. The experimental proof of Huygen's belief came with the YDSE in 1801. 

Young sent coherent light through a pair of vertical slits, and it diffracted into a pattern on the screen kept in front of the slits. The pattern was spread out horizontally, and there were numerous lines on the screen. Without interference and diffraction, light simply would have made two lines on the screen, but this was not the case. This showed that light has a wave nature. 

Experimental Setup of YDSE

The experimental setup of Young’s double-slit experiment is shown below.

Experimental Setup of Young’s Double-Slit Experiment

Light wave theory was better understood through Young's double-slit experiment. A screen or photodetector is located far away from the slits and is denoted by the letter "D." In Young's original double-slit experiment, coherent sources were created by transmitting diffracted light from a single source through two additional slits. At that time coherent sources were not available, but today, we can instead use lasers, which act as coherent sources. One could ask a question about why coherent sources were used for this experiment. Coherent sources were used to maintain a constant phase difference between the two sources of light. 

We will now calculate and derive the results of the YDSE.

Detailed Diagram of YDSE

As we can see from the above diagram, l 1 -l 2, is the path difference between the two waves from the slits. 

Analysis of YDSE

Consider a monochromatic light source called "S" that is separated from two slits by a great distance. S 1 and S 2 are in the same plane as S. S serves as the source for both S 1 and S 2 , making them two reliable coherent sources. We need coherent sources to get the proper interference pattern. 

A screen is located at a distance "D" from the apertures S 1 and S 2 . It receives the light after passing through these slits. The letter "d" stands for the distance between two slits.

It was observed that if any one of the slits was not open, an interference pattern was not formed. So both slits have to be open in order to get interference. The light waves from both slits travel different distances on their way to the screen. From the diagram above, we can see that S 1 P=x and S 2 P=x+$\Delta$x. 

We can represent the light from the two sources as

$\begin{align} &y_{1}=A \sin (k x-\omega t) \\ &y_{2}=B \sin (k(x+\Delta x)-\omega t) \end{align}$

Here, A and B are the amplitudes of the waves, ω is the angular frequency of the wave, and k is the wave number.

Y 2 can be simplified as

$\begin{align} &y_{2}=B \sin (k x-\omega t+k \Delta x) \\ &y_{2}=B \sin (k x-\omega t+\phi) \end{align}$

Here, $\phi$ is the phase difference between the two waves. It can be written as

$\begin{align} &\phi=k \Delta x \\ &\phi=\dfrac{2 \pi}{\lambda} \Delta x \end{align}$

When light interferes, the resultant of this interference can be given by the superposition principle. We have the displacement of the resultant wave as

$\begin{align} &y=y_{1}+y_{2} \\ &y=A \sin (k x-\omega t)+B \sin (k x-\omega t+\phi) \end{align}$

We can further simplify it as

$\begin{align} &y=A \sin (k x-\omega t)+B(\sin (k x-\omega t) \cos \phi+\cos (k x-\omega t) \sin \phi) \\&y=A \sin (\mathrm{kx}-\omega \mathrm{t})+B \sin (\mathrm{kx}-\omega \mathrm{t}) \cos \phi+B \cos (\mathrm{kx}-\omega \mathrm{t}) \sin \phi \\ &y=\sin (\mathrm{kx}-\omega \mathrm{t})(\mathrm{A}+\mathrm{B} \cos \phi)+\cos (\mathrm{kx}-\omega \mathrm{t})(B \sin \phi) \end{align}$

Now, we can simplify this equation further by writing

$\begin{align} &A+B \cos \phi=x \cos \theta \\ &B \sin \phi=x \sin \theta \end{align}$

If we substitute these in the equation before, we get

$\begin{align} &y=x \sin (k x-\omega t) \cos \theta+x \cos (k x-\omega t) \sin \theta \\ &y=x(\sin (k x-\omega t) \cos \theta+\cos (k x-\omega t) \sin \theta) \\ &y=x \sin (k x-\omega t+\theta) \end{align}$

So we find that the resultant wave is indeed a sine wave that has an amplitude x and has a phase difference of $\theta$ from the first wave. 

Now, to find the amplitude x of the resultant wave, we can square both sides and then add the equations

$\begin{align} &(x \cos \theta)^{2}+(x \sin \theta)^{2}=(A+B \cos \phi)^{2}+(B \sin \phi)^{2} \\&x^{2} \cos ^{2} \theta+x^{2} \sin ^{2} \theta=A^{2}+2 A B \cos \phi+B^{2} \cos ^{2} \phi+B^{2} \sin ^{2} \phi \\ &x^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=A^{2}+2 A B \cos \phi+B 2\left(\cos ^{2} \phi+\sin ^{2} \phi\right) \\&x^{2}=A^{2}+B^{2}+2 A B \cos \phi \\ &x=\sqrt{A^{2}+B^{2}+2 A B \cos \phi} \end{align}$

This is the amplitude of the resultant wave, and it is dependent on the amplitudes of the individual waves and a cross term with the cosine of the phase difference between them. 

Constructive and Destructive Interference

We have already said before that we get an interference pattern on the screen. This interference pattern consists of alternate dark and bright spots of light at equal distances from the central bright pattern, which we call the central maxima. We can now formulate a mathematical approach for both the constructive and destructive interference conditions.

We have the amplitude of the resultant wave as

$x=\sqrt{A^{2}+B^{2}+2 A B \cos \phi}$

From this equation, it is clear that the maximum and minimum value of x depends on the cross term between a and b. The maximum value will be

$x=\sqrt{A^{2}+B^{2}+2 A B}$

This maximum value comes when 

$\begin{gather} \cos \phi=1 \\ \phi=2 n \pi \end{gather}$

But we know that

$\phi=\dfrac{2 \pi}{\lambda} \Delta x$

Substituting this, we get

$\begin{align} &\dfrac{2 \pi}{\lambda} \Delta x=2 n \pi \\ &\Delta x=n \lambda \end{align}$

So we see that the waves interfere constructively when the path difference is an integral multiple of the wavelength. These are the conditions for constructive interference. 

Similarly, the minimum value is

$x=\sqrt{a^2+b^2-2ab}$

The minimum value comes when

$\begin{align} &\cos \phi=-1 \\ &\phi=(2 n+1) \pi \\ &\dfrac{2 \pi}{\lambda} \Delta x=(2 n+1) \pi \\ &\Delta x=\left(n+\dfrac{1}{2}\right) \lambda \end{align}$

So we can say that the waves interfere destructively if the path difference is equal to an odd half integral multiple of the wavelength. 

These are the conditions for destructive interference.

The formulas in this section are called ydse formulas. 

Fringe Width

Fringe width refers to the distance between two consecutive bright fringes or any two consecutive dark fringes. From the detailed diagram of YDSE, we can see that D or OB is the separation between the slits and the screen. Similarly, d is the separation between the slits. When the experiment is performed the distance between the slits is too small and the screen is also placed at a good distance from the slits. So we can say that, d < < D.

This is obviously an approximation but a useful one. Under this approximation we could say that paths S 1 P and S 2 P are approximately parallel to each other and we could also say that S 1 A is almost perpendicular to S 1 P, S 2 P, and OP. By approximation, we could then say that, $\angle A S_{1} S_{2}=\theta=\angle P O B$

And because D>>d, this angle will be very small and so,

$\tan \theta \approx \sin \theta \approx \theta$

These are small angle approximations. So the path difference can be written as

$\Delta x=P S_{2}-P S_{1} \approx P S_{2}-P A$

We can write this as

$\begin{align} &\Delta x=S_{2} A=d \sin \theta \\ &\Delta x \approx d \tan \theta \\ &\Delta x \approx d \dfrac{y}{D} \end{align}$

Here y is the distance at which the fringes are formed from the point B in the YDSE diagram. Now for bright fringes we know that

$\Delta x=n\lambda$

So, for bright fringes we can write

$\begin{align} &n \lambda=\dfrac{d y}{D} \\ &\dfrac{n \lambda D}{d}=y \end{align}$

Here n is an integer and for different values of n we can write y as

$y=0, \pm \dfrac{D \lambda}{d}, \pm \dfrac{2 D \lambda}{d}, \cdots$

Similarly for dark fringes we can write the distance from the point B as,

$\begin{align} &y=\left(n+\dfrac{1}{2}\right) \dfrac{D \lambda}{d} \\ &y=\pm \dfrac{D \lambda}{2 d}, \pm \dfrac{3 D \lambda}{2 d}, \pm \dfrac{5 D \lambda}{2 d} \end{align}$

Now the distance between any two consecutive fringes can be calculated. Let’s calculate it for bright fringes by taking the first two values. This will give the fringe width w.

$\begin{align} &w=\dfrac{D \lambda}{d}-0 \\ &w=\dfrac{D \lambda}{d} \end{align}$

This will be the same for dark fringes too, let’s show it now. We’ll take the first two values for dark fringes and calculate the distance between these dark fringes.

$\begin{align} &w=\dfrac{3 D \lambda}{2 d}-\dfrac{D \lambda}{2 d} \\ &w=\dfrac{D \lambda}{d} \end{align}$

The fringe width in Young’s double slit experiment is hence given as

$w=\dfrac{D\lambda}{d}$

The fringe width depends on the distance between the slits, the distance between the slits and the screen and the wavelength of light used.

Intensity in YDSE

What is the intensity of light? We will first understand the meaning of intensity. The brightness or quantity of light produced by a particular lighting source is referred to as the light's intensity. It is a way to quantify how much power, weighted by wavelength, a light source emits. There are specialised high and low light intensity fixtures, lamps, and bulbs, and light intensity changes depending on the lamp source. For instance, fluorescent lights are regarded as "cool" or low-intensity light sources, whereas high-intensity discharge lamps emit light with high intensity. The S.I. unit of the intensity of light is Candela.

The intensity of the wave formula is

$I=\dfrac{1}{2} \varepsilon_{0} a^{2}$

Here, $\varepsilon_{0}$ is the permittivity of free space and a is the amplitude of the wave. This equation is also known as the intensity formula. 

We see that the intensity of the light formula is given by the fact that it is directly proportional to the square of the amplitude. In this case, the amplitude of the resultant wave is,

$x=\sqrt{a^{2}+b^{2}+2 a b \cos \phi}$

This means that the intensity of the resultant wave is,

$I \propto x^{2}$

$I \propto a^{2}+b^{2}+2 a b \cos \phi$

$I=a^{2}+b^{2}+2 a b \cos \phi$

We have taken the proportionality constant to be 1 for the sake of convenience.

For maximum intensity we have, 

$\begin{align} &\cos \phi=1 \\ &\phi=2 n \pi \\ &I=a^{2}+a^{2}+2 a^{2} \\ &I=4 a^{2} \end{align}$

For minimum intensity,

$\begin{align} &\cos \phi=-1 \\ &\phi=(2 n+1) \pi \end{align}$

The minimum intensity is,

$\begin{align} &I=2 a^{2}-2 a^{2} \\ &I=0 \end{align}$

These expressions describe the intensity distribution in the interference pattern. In the interference pattern, we can see that the 2a 1 a 2 is simply transferred from minima to maxima points. The net intensity actually remains constant, or we can say that the average intensity remains conserved. 

Young performed the double-slit experiment in 1801, and it proved the wave nature of light which was highly debatable at that time. Newton believed that light was corpuscular in nature, while Huygens believed that light had a wave nature. Young’s experiment was a big break-through experiment in Physics. The light from two coherent light sources was allowed to pass through two slits and interference patterns were observed on the screen. This interference pattern had bright and dark alternate fringes. This was only possible if the light was a wave. The intensity of the light wave is directly proportional to the square of the amplitude of the wave. The maximum intensity of the fringe in the interference pattern is 4a 2 and the minimum intensity was 0.

FAQs on Intensity in Young’s Double Slit Experiment - Important Topic for JEE

1. Who performed the double-slit experiment with electrons?

The nature of electrons has been debated for quite a long time. Richard Feynman suggested a double-slit experiment for electrons. Richard Feynman's double-slit experiment with electrons was carried out in 2012 at the University of Nebraska–Lincoln using new apparatus that permitted control over the transmission of the two slits and the observation of single-electron detection events. An electron gun shot electrons that travelled through one or two slits that were each 62 nm wide and 4 m tall.

2. What do you understand by monochromatic light?

Single-wavelength light sources are known as monochromatic lights, where mono stands for only one and chroma for colour. Monochromatic lights are defined as visible light with a limited range of wavelengths. It has a wavelength that falls within a constrained wavelength range. These lights can be identified by their brightness, colour, propagation direction, and polarisation state. The LASER is a well-known source of monochromatic light. Light Amplification By Stimulated Emission Of Radiation, or LASER, is its abbreviated form.

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Young’s Double Slit Experiment

Optics is the part of material science that concentrates on the conduct and properties of light, incorporating its connections with issues and the development of instruments that utilise or recognize it. Optics as a rule depicts the conduct of apparent, bright, and infrared light. Since light is an electromagnetic wave, different types of electromagnetic radiation, for example, X-beams, microwaves, and radio waves show comparative properties. Most optical marvels can be represented by utilizing the traditional electromagnetic portrayal of light. 

Complete electromagnetic depictions of light are, in any case, frequently hard to apply practically speaking. Down to earth optics is typically done utilizing worked on models. The most widely recognized of these, mathematical optics, regards light as an assortment of beams that move in straight lines and curve when they go through or reflect from surfaces. Actual optics is a more thorough model of light, which incorporates wave impacts, for example, diffraction and obstruction that can’t be represented in mathematical optics. By and large, the beam based model of light was grown first, trailed by the wave model of light. Progress in the electromagnetic hypothesis in the nineteenth century prompted the disclosure that light waves were indeed electromagnetic radiation.  

Young’s Double Slit Experiment

Young’s double-slit experiment employs two coherent light sources separated by a modest distance, generally a few orders of magnitude larger than the wavelength of light. 

Young’s double-slit experiment aided in the understanding of lightwave theory. The slits are separated by a large distance ‘D’ from a screen or photodetector. Young’s original double-slit experiment employed diffracted light from a single source that was then transmitted through two additional slits to serve as coherent sources. In today’s investigations, lasers are frequently employed as coherent sources.

Each source may be thought of as a coherent lightwave source. The waves travel lengths l1 and l2 to generate a path difference of l at any place on the screen at a distance ‘y’ from the centre. At the sources, the point roughly subtends an angle of (since the distance D is large there is only a very small difference of the angles subtended at sources).

Derivation for Young’s Double Slit Experiment

Consider a monochromatic light source ‘S’ that is kept a long way from two slits s 1 and s 2 . S is in the same plane as s 1 and s 2 . Because both s 1 and s 2 are drawn from S, they act as two consistent sources.

The light travels through these slits and lands on a screen that is positioned at a distance ‘D’ from the apertures s 1 and s 2 . The distance between two slits is denoted by the letter ‘d.’

If s 1 is open and s 2 is closed, the screen opposite s 1 is darkened, leaving just the screen opposite s 2 lit. Only when both slits s 1 and s 2 are open can interference patterns form. 

At the point when the cut partition (d) and the screen distance (D) are kept unaltered, to arrive at P the light waves from s 1 and s 2 should travel various distances. It suggests that there is a way contrast in Young’s twofold cut test between the two light waves from s 1 and s 2 .  

Approximations in Young’s twofold cut analysis-  Guess 1: D > d  Since D > d, the two light beams are thought to be equal.   Guess 2: d/λ >> 1 Often, d is a small amount of a millimeter and λ is a negligible portion of a micrometer for noticeable light.  

Under these conditions θ is little, consequently, we can utilize the estimate,

sin θ = tan θ ≈ θ = λ/d

∴ the path difference, 

Δz = λ/d  

This is the path difference between two waves meeting at a point on the screen. Because of this path difference in Young’s twofold cut investigation, a few focuses on the screen are brilliant and a few focuses are dull.

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, . . . .)

The distance of the nth  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, . . . .)

x = (2n +1)λD/2d

The distance of the nth  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 fringe width is the distance between two consecutive bright (or dark) fringes. β = λD/d

When Young’s double-slit experiment equipment is submerged in a liquid with a refractive index of (μ), the wavelength of light and the fringe width both fall by ” times.

When white light is utilised instead of monochromatic light, coloured fringes appear on the screen, with red fringes being bigger than violet fringes.

Maximum Order of Interference Fringes

On the screen, the position of n th order maximum is,

where n=0, ±1, ±2, …

However, because the 2 nd approximation would be violated, ‘n’ values cannot take infinitely high values. i.e. θ is small (or) y << D.

⇒ γ/D = nλ/d <<1

As a result, the above formula for interference maxima can be used n<< d/λ. When the value of ‘n’ equals that of d/λ, the path difference can no longer be calculated as dγ/D.

Hence for maxima, path difference = nλ

⇒ dsinθ = nλ

​n = dsinθ/λ 

n max = d/λ

The above represents the box function or greatest integer function. Similarly, the highest order of interference minima is given by,

​n min =[d/λ+1/2]

Shape of Interference Fringes in YDSE

The route difference between the two slits is represented by the YDSE diagram.

s 2 p−s 1 p≈dsinθ (constant)

The preceding equation depicts a hyperbola with two foci denoted by the letters s 1 and s 2 .

When we rotate a hyperbola on the axis s 1 s 2 , we obtain an interference pattern on the screen that is a segment of a hyperbola. Fringes are hyperbolic with a straight middle portion of the screen is in the yz plane.

Intensity of Fringes in YDSE

The resulting intensity at location p for two coherent sources s1 and s2 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

  = 2I 0 + 2 (I 0 ) cos φ

  = 2I 0 (1 +  cos φ)

  = 4I 0 cos 2 (ϕ/2)

Constructive and Destructive Interference

The path difference must be an integral multiple of the wavelength for constructive interference to occur. As a result, if a brilliant fringe is at ‘y,’

where n = ±0,1,2,3…..

The centre brilliant fringe is represented by the 0th fringe. Similarly, in Young’s double-slit experiment, the expression for a black fringe may be obtained by setting the path difference to:

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 breakthrough point in science because it proved beyond a shadow of a doubt that light behaved like a wave. Later, the Double Slit Experiment was repeated using electrons, and to everyone’s amazement, the pattern produced was nearly identical to that seen with light. This would forever alter our perceptions of matter and particles, requiring us to believe that matter, like light, acts like a wave.

Sample Questions

Question 1: In Young’s double-slit experiment, the ratio of brightness at minima to peaks is 9:25. Calculate the ratio of the two-slit widths.

I max /I min = (a 1 +a 2 ) 2 /(a 1 -a 2 ) 2                               = 9/25 Solving, a 1 /a 2 = 4/1  Therefore, Ratio of slit width , w 1 /w 2 = a 1 2 /a 2 2 = 16.

Question 2: The distance between the slits in a double-slit experiment is 3 mm, and the slits are 2 m apart from the screen. On the screen, two interference patterns can be observed, one caused by light with a wavelength of 480 nm and the other by light with a wavelength of 600 nm. What is the distance between the fifth-order brilliant fringes of the two interference patterns on the screen?

Separation is given by, y= nλD/d ​where, d = 3 mm = 3 × 10 −3 m D = 2 m λ 1 = 480 nm = 480×10 −9 m λ 2 = 600 nm = 600×10 −9 m n 1 ​=n 2 =5 So, y 1​ =  nλ 1 D/d y 1 = 5×480×10 −9 ×2 / 3×10 −3 ​y 1 =1.6×10 −3 m Also,   y 2 = nλ 2 D /d ​y 2 =  5×600×10 −9 ×2 / 3×10 −3 ​y 2 =2×10 −3 m As , y 2 > y 1 ​y 2 − y 1 ​= 2×10 −3 −1.6×10 −3             = 4×10 −4 m Therefore the separation on the screen between the fifth order bright fringes of the two interference patterns is 4×10 −4 m.

Question 3: The widths of two slits in Young’s experiment are 1: 25. The intensity ratio at the interference pattern’s peaks and minima, I max /I min is

Intensity is proportional to width of the slit. thus, I 1 /I 2 = 1/25 or a 1 /a 2 =  1/5 Imax/Imin = (a 1 +a 2 ) 2 /(a 1 -a 2 ) 2                  = (a 1 +5a 1 ) 2 /(a 1 -5a 1 ) 2                  = 36/16                  = 9/4

Question 4: Two lucid point sources S 1 and S 2 vibrating in stage radiate light of frequency λ. The division between the sources is 2λ. Consider a line going through S 2 and opposite to the line S 1 S 2 . What is the littlest separation from S 2 where at least power happens?

For minimum  intensity of light, path difference must be equal to path difference for dark e.g Δx=λ/2,3λ/2,5λ/2 Consider Δx=(2λ−λ/2)=3λ/2 Now, path difference=S 1 P−S 2 P Δx=  √(4λ 2 +x 2 )−x 3λ/2+x=  √4λ 2 +x 2  Taking square on both sides, 9λ 2 /4+x 2 +3λx=4λ 2 +x 2  3λx=4λ 2 −9λ 2 /4 x=7λ/12 The smallest distance from S 2 where a minimum of intensity occurs is 7λ/12

Question 5: A glass lens is coated with a thin layer having a refractive index of 1.50. The glass has a refractive index of 1.60. What is the thinnest film coating that will reflect 546 nm light extremely strongly (constructive thin film interference)?

Light is entering in a medium of higher refractive index (n 2 =1.60) from a medium of lower index (n 1 =1.50), therefore net path difference in the reflected rays from the two interfaces will be equal to 2t (where t is thickness of thin film) , because when light goes from air to thin film path shift in reflected ray =λ’/2, when light goes from  thin film to glass path shift in reflected ray =λ’/2+2t for constructive interference,     2t=mλ’ or 2t=mλ/n 1 where λ’= wavelength in thin film  , λ=546nm(given) wavelength in vacuum , for thinnest coating , m=0(minimum) , therefore, t min = λ/2n 1                      ​= 546/(2×1.50) = 182nm

Question 6: The initial minimum of the interference pattern of monochromatic light of wavelength e occurs at an angle of λ/a for a single slit of width “a.” We find a maximum for two thin slits separated by a distance “a” at the same angle of λ/a. Explain.

Width of the slit is a. The path difference between two secondary wavelets is given by, Nλ=asinθ Since, θ is very small, sinθ=θ So, for the first order diffraction n=1, the angle is  λ/a  Now, we know that θ must be very small θ=0 (nearly) because of which the diffraction pattern is minimum. Now for interference case, for two interfering waves of intensity l 1 and l 2 ​we must have two slits separated by a distance. We have the resultant intensity, l=l 1 +l−2+ 2 √l 1 l 2 cosθ ​Since, θ=0 (nearly) corresponding to angle λ/a, so cosθ=1 (nearly) So, l=l 1 +l 2  +2 √ l 1 l 2 cosθ ​l=l 1 +l 2 +2 √ l 1 l 2 cos0 ​l=l 1 +l 2 +2 √ l 1 l 2 ​We see the resultant intensity is sum of the two intensities, so there is a maxima corresponding to the angle  λ/a ​This is why at the same angle λ/a we get a maximum for two narrow slits separated by a distance a.

Question 7: The highest and minimum intensities in an interference pattern created by two coherent sources of light have a 9:1 ratio. what is the brightness of the light used?

Imax /Imin = (a 1 +a 2 ) 2 / (a 1 -a 2 ) 2 9/1 = (a 1 +a 2 ) 2 / (a 1 -a 2 ) 2 a 1 /a 2 = 2/1  a 1 2 / a 2 2 = l 1 / l 2                = 4/1 

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