slit experiment formula

Reset password New user? Sign up

Existing user? Log in

Double-slit Experiment

Already have an account? Log in here.

Plane wave representing a particle passing through two slits, resulting in an interference pattern on a screen some distance away from the slits. [1] .

The double-slit experiment is an experiment in quantum mechanics and optics demonstrating the wave-particle duality of electrons , photons , and other fundamental objects in physics. When streams of particles such as electrons or photons pass through two narrow adjacent slits to hit a detector screen on the other side, they don't form clusters based on whether they passed through one slit or the other. Instead, they interfere: simultaneously passing through both slits, and producing a pattern of interference bands on the screen. This phenomenon occurs even if the particles are fired one at a time, showing that the particles demonstrate some wave behavior by interfering with themselves as if they were a wave passing through both slits.

Niels Bohr proposed the idea of wave-particle duality to explain the results of the double-slit experiment. The idea is that all fundamental particles behave in some ways like waves and in other ways like particles, depending on what properties are being observed. These insights led to the development of quantum mechanics and quantum field theory , the current basis behind the Standard Model of particle physics , which is our most accurate understanding of how particles work.

The original double-slit experiment was performed using light/photons around the turn of the nineteenth century by Thomas Young, so the original experiment is often called Young's double-slit experiment. The idea of using particles other than photons in the experiment did not come until after the ideas of de Broglie and the advent of quantum mechanics, when it was proposed that fundamental particles might also behave as waves with characteristic wavelengths depending on their momenta. The single-electron version of the experiment was in fact not performed until 1974. A more recent version of the experiment successfully demonstrating wave-particle duality used buckminsterfullerene or buckyballs , the \(C_{60}\) allotrope of carbon.

Waves vs. Particles

Double-slit experiment with electrons, modeling the double-slit experiment.

To understand why the double-slit experiment is important, it is useful to understand the strong distinctions between wave and particles that make wave-particle duality so intriguing.

Waves describe oscillating values of a physical quantity that obey the wave equation . They are usually described by sums of sine and cosine functions, since any periodic (oscillating) function may be decomposed into a Fourier series . When two waves pass through each other, the resulting wave is the sum of the two original waves. This is called a superposition since the waves are placed ("-position") on top of each other ("super-"). Superposition is one of the most fundamental principles of quantum mechanics. A general quantum system need not be in one state or another but can reside in a superposition of two where there is some probability of measuring the quantum wavefunction in one state or another.

Left: example of superposed waves constructively interfering. Right: superposed waves destructively interfering. [2]

If one wave is \(A(x) = \sin (2x)\) and the other is \(B(x) = \sin (2x)\), then they add together to make \(A + B = 2 \sin (2x)\). The addition of two waves to form a wave of larger amplitude is in general known as constructive interference since the interference results in a larger wave.

If one wave is \(A(x) = \sin (2x)\) and the other is \(B(x) = \sin (2x + \pi)\), then they add together to make \(A + B = 0\) \(\big(\)since \(\sin (2x + \pi) = - \sin (2x)\big).\) This is known as destructive interference in general, when adding two waves results in a wave of smaller amplitude. See the figure above for examples of both constructive and destructive interference.

Two speakers are generating sounds with the same phase, amplitude, and wavelength. The two sound waves can make constructive interference, as above left. Or they can make destructive interference, as above right. If we want to find out the exact position where the two sounds make destructive interference, which of the following do we need to know?

a) the wavelength of the sound waves b) the distances from the two speakers c) the speed of sound generated by the two speakers

This wave behavior is quite unlike the behavior of particles. Classically, particles are objects with a single definite position and a single definite momentum. Particles do not make interference patterns with other particles in detectors whether or not they pass through slits. They only interact by colliding elastically , i.e., via electromagnetic forces at short distances. Before the discovery of quantum mechanics, it was assumed that waves and particles were two distinct models for objects, and that any real physical thing could only be described as a particle or as a wave, but not both.

In the more modern version of the double slit experiment using electrons, electrons with the same momentum are shot from an "electron gun" like the ones inside CRT televisions towards a screen with two slits in it. After each electron goes through one of the slits, it is observed hitting a single point on a detecting screen at an apparently random location. As more and more electrons pass through, one at a time, they form an overall pattern of light and dark interference bands. If each electron was truly just a point particle, then there would only be two clusters of observations: one for the electrons passing through the left slit, and one for the right. However, if electrons are made of waves, they interfere with themselves and pass through both slits simultaneously. Indeed, this is what is observed when the double-slit experiment is performed using electrons. It must therefore be true that the electron is interfering with itself since each electron was only sent through one at a time—there were no other electrons to interfere with it!

When the double-slit experiment is performed using electrons instead of photons, the relevant wavelength is the de Broglie wavelength \(\lambda:\)

\[\lambda = \frac{h}{p},\]

where \(h\) is Planck's constant and \(p\) is the electron's momentum.

Calculate the de Broglie wavelength of an electron moving with velocity \(1.0 \times 10^{7} \text{ m/s}.\)

Usain Bolt, the world champion sprinter, hit a top speed of 27.79 miles per hour at the Olympics. If he has a mass of 94 kg, what was his de Broglie wavelength?

Express your answer as an order of magnitude in units of the Bohr radius \(r_{B} = 5.29 \times 10^{-11} \text{m}\). For instance, if your answer was \(4 \times 10^{-5} r_{B}\), your should give \(-5.\)

Image Credit: Flickr drcliffordchoi.

While the de Broglie relation was postulated for massive matter, the equation applies equally well to light. Given light of a certain wavelength, the momentum and energy of that light can be found using de Broglie's formula. This generalizes the naive formula \(p = m v\), which can't be applied to light since light has no mass and always moves at a constant velocity of \(c\) regardless of wavelength.

The below is reproduced from the Amplitude, Frequency, Wave Number, Phase Shift wiki.

In Young's double-slit experiment, photons corresponding to light of wavelength \(\lambda\) are fired at a barrier with two thin slits separated by a distance \(d,\) as shown in the diagram below. After passing through the slits, they hit a screen at a distance of \(D\) away with \(D \gg d,\) and the point of impact is measured. Remarkably, both the experiment and theory of quantum mechanics predict that the number of photons measured at each point along the screen follows a complicated series of peaks and troughs called an interference pattern as below. The photons must exhibit the wave behavior of a relative phase shift somehow to be responsible for this phenomenon. Below, the condition for which maxima of the interference pattern occur on the screen is derived.

Left: actual experimental two-slit interference pattern of photons, exhibiting many small peaks and troughs. Right: schematic diagram of the experiment as described above. [3]

Since \(D \gg d\), the angle from each of the slits is approximately the same and equal to \(\theta\). If \(y\) is the vertical displacement to an interference peak from the midpoint between the slits, it is therefore true that

\[D\tan \theta \approx D\sin \theta \approx D\theta = y.\]

Furthermore, there is a path difference \(\Delta L\) between the two slits and the interference peak. Light from the lower slit must travel \(\Delta L\) further to reach any particular spot on the screen, as in the diagram below:

Light from the lower slit must travel further to reach the screen at any given point above the midpoint, causing the interference pattern.

The condition for constructive interference is that the path difference \(\Delta L\) is exactly equal to an integer number of wavelengths. The phase shift of light traveling over an integer \(n\) number of wavelengths is exactly \(2\pi n\), which is the same as no phase shift and therefore constructive interference. From the above diagram and basic trigonometry, one can write

\[\Delta L = d\sin \theta \approx d\theta = n\lambda.\]

The first equality is always true; the second is the condition for constructive interference.

Now using \(\theta = \frac{y}{D}\), one can see that the condition for maxima of the interference pattern, corresponding to constructive interference, is

\[n\lambda = \frac{dy}{D},\]

i.e. the maxima occur at the vertical displacements of

\[y = \frac{n\lambda D}{d}.\]

The analogous experimental setup and mathematical modeling using electrons instead of photons is identical except that the de Broglie wavelength of the electrons \(\lambda = \frac{h}{p}\) is used instead of the literal wavelength of light.

• Lookang, . CC-3.0 Licensing . Retrieved from https://commons.wikimedia.org/w/index.php?curid=17014507
• Haade, . CC-3.0 Licensing . Retrieved from https://commons.wikimedia.org/w/index.php?curid=10073387
• Jordgette, . CC-3.0 Licensing . Retrieved from https://commons.wikimedia.org/w/index.php?curid=9529698

Problem Loading...

Note Loading...

Set Loading...

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.

Share This Book

• He placed a screen that had two slits cut into it in front of a monochromatic ( single color ) light.
• The results of Young's Double Slit Experiment should be very different if light is a wave or a particle.
• Let’s look at what the results would be in both situations, and then see how this experiment supports the wave model.

If light is a particle…

We set up our screen and shine a bunch of monochromatic light onto it.

• Imagine it as being almost as though we are spraying paint from a spray can through the openings.
• Since they are little particles they will make a pattern of two exact lines on the viewing screen ( Figure 1 ).

If light is a wave…

If light is a wave, everything starts the same way, but results we get are very different.

• Remember, diffraction is when light passes through a small opening and starts to spread out. This will happen from both openings ( Figure 2 ).
• Where crest meets crest , there will be constructive interference and the waves will make it to the viewing screen as a bright spot .
• Where crest meets trough there will be destructive interference that cancel each other out… a black spot will appear on the screen.
• When this experiment is performed we actually see this, as shown in Figure 3 .

We must conclude that light is made up of waves, since particles can not diffract.

Calculations

When you set up this sort of an apparatus, there is actually a way for you to calculate where the bright lines (called fringes ) will appear.

• There is always a middle line, which is the brightest. We call it the central fringe .
• The central fringe is n = 0 .
• The fringe to either side of the central fringe has an order of n = 1 (the first order fringe ).
• The order of the next fringe out on either side is n = 2 (the second order fringe ).
• And so on, as shown in Figure 4 .

The formula that we will use to figure out problems involving double slit experiments is easy to mix up, so make sure you study it carefully.

λ = wavelength of light used (m) x = distance from central fringe (m) d = distance between the slits (m) n = the order of the fringe L = length from the screen with slits to the viewing screen (m)

It is very easy to mix up the measurements of x, d, and L.

• Make sure to look at Figure 5 and see the different things each is measuring.
• If you mix up x and d it's not so bad, since they are both on top in the formula. If you were to mix them up with L, you would get the wrong answer.
• Almost all questions that you will see for this formula just involve sorting out what each variable is... you might find it helpful to write out a list of givens.

Example 1 : A pair of screens are placed 13.7m apart. A third order fringe is seen on the screen 2.50cm from the central fringe. If the slits were cut 0.0960 cm apart, determine the wavelength of this light. Roughly what colour is it?

Just to make sure you’ve got all the numbers from the question matched with the correct variables… L = 13.7 m n = 3 x = 2.50cm = 0.0250 m d = 0.0960cm = 9.60e-4 m It’s probably a yellow light being used given the wavelength we've measured.

If a white light is used in the double slit experiment, the different colours will be split up on the viewing screen according to their wavelengths.

• The violet end of the spectrum (with the shortest wavelengths) is closer to the central fringe, with the other colours being further away in order.

There is also a version of the formula where you measure the angle between the central fringe and whatever fringe you are measuring.

• The formula works the same way, with the only difference being that we measure the angle instead of x and L.
• Make sure that your calculator is in degree mode before using this version of the formula.

Example 2 : If a yellow light with a wavelength of 540 nm shines on a double slit with the slits cut 0.0100 mm apart, determine what angle you should look away from the central fringe to see the second order fringe?

Do not forget to: Change the wavelength into metres. Change the slit separation into metres. "Second order" is a perfect number and has an infinite number of sig digs.

The Single Slit

A surprising experiment is that you can get the same effect from using a single slit instead of a double slit.

• In Figure 7 , the blue path has to travel further than the red path... if this difference is equal to half a wavelength, they will meet each other out of sync .
• If they meet crest to crest or trough to trough they will be in sync , but if they meet crest to trough they will be out of sync .
• Being in sync will result in constructive interference , while meeting out of sync will result in destructive interference .
• After the first couple of fringes (n = 1 and 2), the edges start getting really fuzzy, so you have a hard time measuring anything.
• The only real difference in calculations is that "d" is now the width of the single opening.
• If two slits work better than one, would more than two slits work better? This is a question that we will answer in the next section.

Example 3 : For a single slit experiment apparatus like the one described above, determine how far from the central fringe the first order violet (λ = 350nm) and red (λ = 700nm) colours will appear if the screen is 10 m away and the slit is 0.050 cm wide.

We need to solve the formula for “x”, the distance from the central fringe. For the violet light… For the red light…

4.3 Double-Slit Diffraction

Learning objectives.

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

• Describe the combined effect of interference and diffraction with two slits, each with finite width
• Determine the relative intensities of interference fringes within a diffraction pattern
• Identify missing orders, if any

When we studied interference in Young’s double-slit experiment, we ignored the diffraction effect in each slit. We assumed that the slits were so narrow that on the screen you saw only the interference of light from just two point sources. If the slit is smaller than the wavelength, then Figure 4.10 (a) shows that there is just a spreading of light and no peaks or troughs on the screen. Therefore, it was reasonable to leave out the diffraction effect in that chapter. However, if you make the slit wider, Figure 4.10 (b) and (c) show that you cannot ignore diffraction. In this section, we study the complications to the double-slit experiment that arise when you also need to take into account the diffraction effect of each slit.

To calculate the diffraction pattern for two (or any number of) slits, we need to generalize the method we just used for a single slit. That is, across each slit, we place a uniform distribution of point sources that radiate Huygens wavelets, and then we sum the wavelets from all the slits. This gives the intensity at any point on the screen. Although the details of that calculation can be complicated, the final result is quite simple:

Two-Slit Diffraction Pattern

The diffraction pattern of two slits of width a that are separated by a distance d is the interference pattern of two point sources separated by d multiplied by the diffraction pattern of a slit of width a .

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 to interference fringes. Equation 4.1 also uses m , but this time to refer to diffraction minima. If both equations are used simultaneously, it is good practice to use a different variable (such as n ) for one of these integers in order to keep them distinct.]

Interference and diffraction effects operate simultaneously and generally produce minima at different angles. This gives rise to a complicated pattern on the screen, in which some of the maxima of interference from the two slits are missing if the maximum of the interference is in the same direction as the minimum of the diffraction. We refer to such a missing peak as a missing order . One example of a diffraction pattern on the screen is shown in Figure 4.11 . The solid line with multiple peaks of various heights is the intensity observed on the screen. It is a product of the interference pattern of waves from separate slits and the diffraction of waves from within one slit.

Example 4.3

Intensity of the fringes.

From Equation 4.4 ,

Substituting from above,

For a = 2 λ a = 2 λ , d = 6 λ d = 6 λ , and m = 1 m = 1 ,

Then, the intensity is

Significance

Example 4.4, two-slit diffraction.

Using d sin θ = m λ d sin θ = m λ for θ = 2.5 × 10 −2 rad θ = 2.5 × 10 −2 rad , we find

which is the maximum interference order that fits inside the central peak. We note that m = ± 10 m = ± 10 are missing orders as θ θ matches exactly. Accordingly, we observe bright fringes for

for a total of 19 bright fringes.

Check Your Understanding 4.3

For the experiment in Example 4.4 , show that m = 20 m = 20 is also a missing order.

Interactive

Explore the effects of double-slit diffraction. In this simulation written by Fu-Kwun Hwang, select N = 2 N = 2 using the slider and see what happens when you control the slit width, slit separation and the wavelength. Can you make an order go “missing?”

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 OpenStax.

Access for free at https://openstax.org/books/university-physics-volume-3/pages/1-introduction

• Authors: Samuel J. Ling, Jeff Sanny, William Moebs
• Publisher/website: OpenStax
• Book title: University Physics Volume 3
• Publication date: Sep 29, 2016
• Location: Houston, Texas
• Book URL: https://openstax.org/books/university-physics-volume-3/pages/1-introduction
• Section URL: https://openstax.org/books/university-physics-volume-3/pages/4-3-double-slit-diffraction

© Jul 23, 2024 OpenStax. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . 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.

• Why Does Water Expand When It Freezes
• Gold Foil Experiment
• Faraday Cage
• Oil Drop Experiment
• Magnetic Monopole
• Why Do Fireflies Light Up
• Types of Blood Cells With Their Structure, and Functions
• The Main Parts of a Plant With Their Functions
• Parts of a Flower With Their Structure and Functions
• Parts of a Leaf With Their Structure and Functions
• Why Does Ice Float on Water
• Why Does Oil Float on Water
• How Do Clouds Form
• What Causes Lightning
• How are Diamonds Made
• Types of Meteorites
• Types of Volcanoes
• Types of Rocks

Double Slit Experiment

What is the double-slit experiment.

The double-slit experiment is a test that demonstrates light can fundamentally display both wave and particle features, also known as wave-particle (photon) duality. It consists of two closely-spaced slits in front of a light source such that a beam of light passing through them is split and projected on a screen placed far from the slits. The split beam forms two coherent sources of light. A pattern of bright and dark fringes appears on the screen due to the interference of light waves coming from these two sources. Interference is the phenomenon in which when waves are superimposed, they may reinforce or cancel each other out. It is most pronounced when the wavelength of the radiation is comparable to the linear dimensions of the slits.

The history of this experiment goes back to 1801 when British polymath Thomas Young demonstrated the wave nature of light and hence, is often called Young’s double-slit experiment.

Double-Slit Experiment Equation

A monochromatic light source falls on a screen that contains two parallel slits, which serve as the sources of coherent light. The light waves emerging from the two slits then interfere and form an interference pattern on the viewing screen. The bright fringes or bands correspond to interference maxima, and the dark fringes are interference minima.

To locate the fringes from the center of the screen, one assumes that the distance between the screen and slit is much greater than the distance separating the slits. Also, the separation between the slits is much greater than the wavelength of the monochromatic light.

By measuring the separation between the fringes and knowing the distance of the screen from the slits and distance between the two slits, it is possible to determine the wavelength of light.

Ans. The double-slit experiment was a groundbreaking test that decisively proves the theory that light behaves as a wave.

Ans . Yes. Many researchers have performed the double-slit experiment on electrons and found similar results to that of photons.

Article was last reviewed on Wednesday, June 3, 2020

Related articles

Leave a Reply Cancel reply

Your email address will not be published. Required fields are marked *

Save my name, email, and website in this browser for the next time I comment.

Popular Articles

Join our Newsletter

Fill your E-mail Address

Related Worksheets

• Privacy Policy

© 2024 ( Science Facts ). All rights reserved. Reproduction in whole or in part without permission is prohibited.

Young's double slit experiment

This experiment to use the effects of interference to measure the wavelength of light was devised by Thomas Young in 1801, although the original idea was due to Grimaldi. The method produces non-localised interference fringes by division of wavefront, and a sketch of the experimental arrangement is shown in Figure 1.

Light from a monochromatic line source passes through a lens and is focused on to a single slit S. It then falls on a double slit (S 1 and S 2 ) and this produces two wave trains that interfere with each other in the region on the right of the diagram. The interference pattern at any distance from the double slit may be observed with a micrometer eyepiece or by placing a screen in the path of the waves. The separation across double slit should be less than 1 mm, the width of each slit about 0.3 mm, and the distance between the double slit and the screen between 50 cm and 1 m. The single slit, the source and the double slit must be parallel to produce the optimum interference pattern. Alternatively a laser may be used and the fringes viewed on a screen some metres away without the need for a micrometer eyepiece or a single slit. The formula relating the dimensions of the apparatus and the wavelength of light may be proved as follows.

Note that the fringe width is directly proportional to the wavelength, and so light with a longer wavelength will give wider fringes. Although the diagram shows distinct light and dark fringes, the intensity actually varies as the cos 2 of angle from the centre. If white light is used a white centre fringe is observed, but all the other fringes have coloured edges, the blue edge being nearer the centre. Eventually the fringes overlap and a uniform white light is produced.

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

• 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-

Frequently Asked Questions – FAQs

What is meant by interference, name a coherent source of light., which type of fringes are produced from the destructive interference, in which type of interference, the waves will have the same displacement and phase., does the young’s double slit experiment prove the interference of light.

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 Physics 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

loads of thanks for providing these free study materials will help students a lot

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.