dark fringe experiment

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Young's Double Slit Experiment

What is young’s double slit experiment.

Young’s double slit experiment uses two coherent sources of light placed at a small distance apart. Usually, only a few orders of magnitude greater than the wavelength of light are used. Young’s double slit experiment helped in understanding the wave theory of light , which is explained with the help of a diagram. As shown, a screen or photodetector is placed at a large distance, ‘D’, away from the slits.

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The original Young’s double slit experiment used diffracted light from a single source passed into two more slits to be used as coherent sources. Lasers are commonly used as coherent sources in modern-day experiments.

Table of Contents

• Position of Fringes
• Shape of Fringes
• Intensity of Fringes

Special Cases

• Displacement of Fringes

Each source can be considered a source of coherent light waves . At any point on the screen at a distance ‘y’ from the centre, the waves travel distances  l 1  and  l 2 to create a path difference of Δl at the point. The point approximately subtends an angle of θ at the sources (since the distance D is large, there is only a very small difference between the angles subtended at sources).

Derivation of Young’s Double Slit Experiment

Consider a monochromatic light source ‘S’ kept at a considerable distance from two slits: s 1 and s 2 . S is equidistant from s 1 and s 2 . s 1 and s 2 behave as two coherent sources as both are derived from S.

The light passes through these slits and falls on a screen which is at a distance ‘D’ from the position of slits s 1 and s 2 . ‘d’ is the separation between two slits.

If s 1 is open and s 2 is closed, the screen opposite to s 1 is closed, and only the screen opposite to s 2 is illuminated. The interference patterns appear only when both slits s 1 and s 2 are open.

When the slit separation (d) and the screen distance (D) are kept unchanged, to reach P, the light waves from s 1 and s 2 must travel different distances. It implies that there is a path difference in Young’s double slit experiment between the two light waves from s 1 and s 2 .

Approximations in Young’s double slit experiment 

• Approximation  1: D > > d: Since D > > d, the two light rays are assumed to be parallel.
• Approximation  2:   d/λ >> 1:  Often, d is a fraction of a millimetre, and λ is a fraction of a micrometre for visible light.

Under these conditions, θ is small. Thus, we can use the approximation sin θ = tan θ ≈ θ = λ/d.

∴ path difference, Δz = λ/d

This is the path difference between two waves meeting at a point on the screen. Due to this path difference in Young’s double slit experiment, some points on the screen are bright, and some points are dark.

Now, we will discuss the position of these light and dark fringes and fringe width.

Position of Fringes in Young’s Double Slit Experiment

Position of bright fringes.

For maximum intensity or bright fringe to be formed at P,

Path difference, Δz = nλ (n = 0, ±1, ±2, . . . .)

i.e., xd/D = nλ

The distance of the n th   bright fringe from the centre is

x n = nλD/d

Similarly, the distance of the (n-1) th  bright fringe from the centre is

x (n-1) = (n -1)λD/d

Fringe width,  β =  x n   – x (n-1)  = nλD/d – (n -1)λD/d = λD/d

(n = 0, ±1, ±2, . . . .)

Position of Dark Fringes

For minimum intensity or dark fringe to be formed at P,

Path difference, Δz = (2n + 1) (λ/2) (n = 0, ±1, ±2, . . . .)

i.e., x = (2n +1)λD/2d

The distance of the n th   dark fringe from the centre is

x n = (2n+1)λD/2d

x (n-1) = (2(n-1) +1)λD/2d

Fringe width, β = x n   – x (n-1)  = (2n + 1) λD/2d – (2(n -1) + 1)λD/2d = λD/d

Fringe Width

The distance between two adjacent bright (or dark) fringes is called the fringe width.

If the apparatus of Young’s double slit experiment is immersed in a liquid of refractive index  (μ), then the wavelength of light and fringe width decreases ‘μ’ times.

If white light is used in place of monochromatic light, then coloured fringes are obtained on the screen, with red fringes larger in size than violet.

Angular Width of Fringes

Let the angular position of n th bright fringe is θ n, and because of its small value, tan θ n   ≈ θ n

Similarly, the angular position of (n+1) th bright fringe is θ n+1,  then

∴ The angular width of a fringe in Young’s double slit experiment is given by,

Angular width is independent of ‘n’, i.e., the angular width of all fringes is the same.

Maximum Order of Interference Fringes

But ‘n’ values cannot take infinitely large values as it would violate the 2 nd approximation.

i.e., θ is small (or) y < < D

When the ‘n’ value becomes comparable to d/ λ, path difference can no longer be given by d γ/D.

Hence for maxima, path difference = nλ

The above represents the box function or greatest integer function.

Similarly, the highest order of interference minima

The Shape of Interference Fringes in YDSE

From the given YDSE diagram, the path difference between the two slits is given by

The above equation represents a hyperbola with its two foci as, s 1 and s 2 .

The interference pattern we get on the screen is a section of a hyperbola when we revolve the hyperbola about the axis s 1 s 2 .

If the screen is a yz plane, fringes are hyperbolic with a straight central section.

If the screen is xy plane , the fringes are hyperbolic with a straight central section.

The Intensity of Fringes in Young’s Double Slit Experiment

For two coherent sources, s 1 and s 2 , the resultant intensity at point p is given by

I = I 1 + I 2 + 2 √(I 1 . I 2 ) cos φ

Putting I 1 = I 2 = I 0  (Since, d<<<D)

I = I 0 + I 0 + 2 √(I 0 .I 0 ) cos φ

I = 2I 0 + 2 (I 0 ) cos φ

I = 2I 0 (1 +  cos φ)

For maximum intensity

phase difference   φ = 2nπ

Then, path difference \(\begin{array}{l}\Delta x=\frac{\lambda }{{2}{\pi }}\left( {2}n{\pi } \right)\end{array} \) = nλ

The intensity of bright points is maximum and given by

I max = 4I 0

For minimum intensity

φ = (2n – 1) π

Phase difference φ = (2n – 1)π

Thus, the intensity of minima is given by

If I 1 ≠ I 2 , I min ≠ 0.

Rays Not Parallel to Principal Axis:

From the above diagram,

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

Source Placed beyond the Central Line:

If the source is placed a little above or below this centre line, the wave interaction with S 1 and S 2 has a path difference at point P on the screen.

Δ x= (distance of ray 2) – (distance of ray 1)

= bd/a + yd/D → (*)

We know Δx = nλ for maximum

Δx = (2n – 1) λ/2 for minimum

By knowing the value of Δx from (*), we can calculate different positions of maxima and minima .

Displacement of Fringes in YDSE

When a thin transparent plate of thickness ‘t’ is introduced in front of one of the slits in Young’s double slit experiment, the fringe pattern shifts toward the side where the plate is present.

The dotted lines denote the path of the light before introducing the transparent plate. The solid lines denote the path of the light after introducing a transparent plate.

Where μt is the optical path.

Then, we get,

Term (1) defines the position of a bright or dark fringe; term (2) defines the shift that occurred in the particular fringe due to the introduction of a transparent plate.

Constructive and Destructive Interference

For constructive interference, the path difference must be an integral multiple of the wavelength.

Thus, for a bright fringe to be at ‘y’,

Or, y = nλD/d

Where n = ±0,1,2,3…..

The 0th fringe represents the central bright fringe.

Similarly, the expression for a dark fringe in Young’s double slit experiment can be found by setting the path difference as

Δl = (2n+1)λ/2

This simplifies to

(2n+1)λ/2 = y d/D

y = (2n+1)λD/2d

Young’s double slit experiment was a watershed moment in scientific history because it firmly established that light behaved like a wave.

The double slit experiment was later conducted using electrons , and to everyone’s surprise, the pattern generated was similar as expected with light. This would forever change our understanding of matter and particles, forcing us to accept that matter, like light, also behaves like a wave.

Wave Optics

Young’s double slit experiment.

Frequently Asked Questions on Young’s Double Slit Experiment

What was the concept explained by young’s double slit experiment.

Young’s double slit experiment helps in understanding the wave theory of light.

What are the formulas derived from Young’s double slit experiment?

For constructive interference, dsinθ = mλ , for m = 0,1,-1,2,-2

For destructive interference, dsinθ = (m+½)λ, for m = 0,1,-1,2,-2 Here, d is the distance between the slits. λ is the wavelength of the light waves.

What is called a fringe width?

The distance between consecutive bright or dark fringe is called the fringe width.

What kind of source is used in Young’s double slit experiment?

A coherent source is used in Young’s double slit experiment.

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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

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.

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

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

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

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

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

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

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 ).

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.

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?

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

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

Example 1: Finding a Wavelength from an Interference Pattern

Substituting known values yields

Example 2: Calculating Highest Order Possible

Strategy and Concept

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.

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.

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

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?

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 ).

9: 1200 nm (not visible)

11: (a) 760 nm

(b) 1520 nm

For two adjacent fringes we have,

Subtracting these equations gives

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Order of a dark fringe in Young's double slit experiment

• Thread starter FranzDiCoccio
• Start date Apr 10, 2020
• Tags Double slit Double slit experiment Experiment Fringe Slit Young's double slit
• Apr 10, 2020
• Findings from experimental setup demonstrate potential for compact and portable nuclear clocks
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A PF Singularity

Is there a reason why you don't think m=0 exists for the dark fringes? If you look at the equation, for m=0, you have a non-zero angle on either side of the central bright fringe. So for the bright fringe, m=0 is at θ = 0, but for the dark fringe, m=0 is at a non-zero angle. Besides, this is nothing more than a "counting index". Zz.  

• Apr 11, 2020

ZapperZ said: Is there a reason why you don't think m=0 exists for the dark fringes?

ZapperZ said: Besides, this is nothing more than a "counting index".

FranzDiCoccio said: but it might very well be personal taste.

Hi sophiecentaur, thanks for your reply. I surely agree with you. If you're interacting with a human, you can actually explain yourself and agree on a notation. I guess that, whenever I can, I'll refrain from using the term "order" and say something like "the nth (bright/dark) fringe to either side of the central one". This is unambiguous. However, suppose that I find a question (e.g. in a test) asking for the angular position of the second-order dark fringe. Is this the second from the central one, or is it the third ? If you start counting from zero, it would be the third. A common notation should be established for that question to be unambiguous. By the way, the formula for a single slit is [tex] d \sin \theta = m \lambda, \qquad m = \pm 1, \pm 2,\ldots [/tex] Hence, in that case, there seems to be no problem in starting from 1. It seems that the first-order dark fringe is the first one on the side of the central bright fringe for a single slit, but the second one for a double slit? Very confusing. My feeling is: order primarily refers to bright fringes, which are the most visibile ones by definition. I guess that the concept has been "extended" to dark fringes in (some) textbooks, mainly for having some assortment in exercises. I'm not sure dark fringes are ever used in actual experiments. This generalization has been a bit too casual, and someone started using 2m+1 for odd numbers, which forces you to start counting from 0. I have the feeling that "order" mainly applies to bright fringes, where there is no ambiguity, and one should be careful in extending this concept to dark fringes. I'm thinking of a diffraction grating, e.g.  

FAQ: Order of a dark fringe in Young's double slit experiment

1. what is the order of a dark fringe in young's double slit experiment.

The order of a dark fringe in Young's double slit experiment refers to the number assigned to a particular dark fringe, starting from the central bright fringe. It is used to describe the location of the fringe and can be calculated using the equation m = (x - x 0 )/λL, where m is the order, x is the distance from the central bright fringe, x 0 is the distance between the two slits, λ is the wavelength of light, and L is the distance from the slits to the screen.

2. How is the order of a dark fringe determined in Young's double slit experiment?

The order of a dark fringe can be determined by measuring the distance from the central bright fringe to the dark fringe and using the equation m = (x - x 0 )/λL. This equation takes into account the distance between the slits, the wavelength of light, and the distance from the slits to the screen. The resulting value of m will correspond to the order of the dark fringe.

3. What does the order of a dark fringe indicate in Young's double slit experiment?

The order of a dark fringe indicates the location of the dark fringe in relation to the central bright fringe. A higher order indicates a darker fringe that is further away from the central bright fringe, while a lower order indicates a lighter fringe that is closer to the central bright fringe.

4. Can the order of a dark fringe change in Young's double slit experiment?

Yes, the order of a dark fringe can change depending on the parameters of the experiment. For example, if the distance between the slits or the distance from the slits to the screen is changed, the order of the dark fringe will also change. Additionally, the order of the dark fringe can be affected by the wavelength of light used in the experiment.

5. How does the order of a dark fringe affect the interference pattern in Young's double slit experiment?

The order of a dark fringe affects the interference pattern by determining the location and intensity of the dark fringes. The higher the order of the dark fringe, the further away it is from the central bright fringe and the darker it will appear. This creates a distinct pattern of light and dark fringes on the screen, known as an interference pattern, which is a result of the interference of light waves from the two slits.

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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?”

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• Why are the dark fringes dark in the double slit experiment?

There are a lot of questions about the double slit experiment on this site, none of them answer my question specifically, and there are basically two theories why the dark fringes should be dark:

• every single photon, that is shot one at a time, lands on the bright area, thus, no photon lands in the dark area, that is clearly understandable

All photons passing thru the slits leave a dot on the screen, this is true for single or multiple photon intensities

Why does the photon strike at one or another place on the tape?

In principle every photon that passes the double slit should end on the screen.

In a double slit experiment, does each and every photon leave a dot on the screen in the bright area?

• photons go everywhere, but energy from the dark fringes returns to the apparatus

It experimentally shows that the light from the dark fringes goes back to the source of the collimated beams.

Destructive interference

For so long, I thought that 1. was the true explanation. Somehow I have a hard time reconciling both 1. and 2. together. If every single photon that is shot, leaves a bright spot, then all the energy from the photons is absorbed there, locally. There is no energy going to the dark areas at all. Then what energy is returning to the apparatus?

• quantum-mechanics
• electromagnetism
• double-slit-experiment

4 Answers 4

Each photon that goes in has to end up somewhere! In the Young's slits case, the photons don't go to the dark areas and go the bright areas instead. So that's where they take their energy with them. There are other interferometer configurations where the light exits the interferometer in other directions. For example, in the Michelson, the light either comes out in the usual output direction, or its gets reflected back towards the source. You can check that the layout of the mirrors will produce exactly this effect.

The dark fringes are dark because no energy is delivered there. The energy does not first go to a dark area and then go away to somewhere else. The energy simply does not arrive at the dark area at all. "Energy" and "photons" are two ways of talking about the same thing.

Dark areas in light interference in general can be assigned either to your 1. or 2.

• is for simple diffractive interference of electromagnetic waves with matter, as the double slit experiment. Basically the scattering of photons on fringe electric fields of matter with given boundary conditions.

But dark fringes can appear with interference of two electromagnetic beams, and in that case the video linked shows that there is return to the source , which is your 2.

So one has to be careful in assigning reasons for interference patterns in general.

• $\begingroup$ Let me add that in the typical double-slit experiment the missing energy in the dark fringes doesn't go back to the source, it's only redistributed – namely into the bright fringes. In the linked video about the Michelson interferometer the geometry is arranged differently. There, the missing energy is ideed reflected back to the source. $\endgroup$ –  A. P. Commented Nov 29, 2020 at 22:44

Young concluded about light as a wave from the fringes behind edges and was satisfied. Newton was angry that his corpuscle theory was no longer correct but was not able to get a good reason for the fringes. And Einstein rehabilitated Newton with the introduction of quanta of light.

There can be no doubt that the following statements on the nature of EM radiation are clear:

• EM radiation always starts with the emission of photons from excited particles
• during their life the photon is a indivisible quanta
• the photon has a magnetic and an electric field component, both with an oscillating intensity

Is there an interaction between the photon and the edge it flies past an edge?

I never understood why the such question is not asked in our days.

The interference behind edges is out since a long time. We know, that photons interact only very weak for the low intensities we are using for our experiments.

The intensity distribution (fringes) appear not only behind double (or more or one) slits, but also behind edges. And the same result (statistically) you get, fire the photons one by one.

A good approach for the possible rehabilitation of Newton is the investigation of the interaction between the photons electric and magnetic field and the surface electrons of an edge.

To become friends with this idea one could remember that a polarizer grid (designed for some wavelengths) rotates some portion of the not polarized light. This can be seen by the experiment where one put a third polarizer with 45° between two with 0° and 90° and the totally blocked light behind the 0° and 90° polarizers is no longer blocked by the help of the 45° polarizer.

Experiments with changing electrical potential on the edge or in an external magnetic field could confirm or reject the relevance of the approach. For electrons behind edges such experiments was carried out by G. Möllenstedt (not available in the English Wikipedia) and H. Düker in a biprisma experiment:

The distances of the fringes was a function of the electric potential of the wire.

A metallic mirror can be seen an example of case 2. consider a mirror hit be a perpendicular plane wave, say along the +z direction. Electrons in the metal move out of phase with incoming light and emit an out of phase wave in the +z and in the -z direction. The incoming wave and the out of phase wave along +z cancel out due to destructive interference. What is left is an out of phase wave in the -z direction, that is, a reflected wave.

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Young's Double Slit Experiment Derivation

• Youngs Double Slit Experiment Derivation

Learn about Young's Double Slit Experiment Derivation

In 1801, an English physician and physicist established the principle of light interference, where he made a pinhole camera in cardboard and allowed sunlight to pass through it.

This white light was then allowed to fall upon another cardboard having two pinholes placed together symmetrically. 

The emerging light was received on a plane screen placed at some distance. 

At a given point on the screen, the waves emerging from two holes had different phases, interfering to give a pattern of bright and dark areas.

Such a variation of intensity on the plane screen demonstrated the light waves emerging from the two holes.

Therefore, this pattern of bright (constructive fringe) and dark (destructive fringe) areas can be sharply defined only if the light of a single wavelength is used.

(Image to be added soon)

Young Double Slits Experiment Derivation

The two waves interfering at P have covered different distances.

Let the waves from two coherent sources of light be represented as 

                  y 1 = a sinωt…(1)

                  y 2 = b Sin(ωt + Φ)...(2)

Here, a and b are amplitudes of the two waves resp. 

Φ is the constant phase angle by which the second wave leads the first wave.

Applying the superposition principle, the displacement(y) of the resultant wave at time (t) would be:

 y  = y 1 +  y 2 =   a sinωt +  b sin(ωt + Φ)

                        Expanding sin(ωt + Φ) = sin ωt cosΦ + cosωt . sinΦ

                          =    sinωt  (a+ b cosΦ ) + cosωt . b sinΦ

Put = a + b cosΦ = R cosӨ ….(3)

and b sinΦ = R sinӨ ….(4)

y  = sinωt. R cosӨ + cosωt . R sinӨ 

             y  = R sin(ωt + Ө)......(4)

The wave equation (4) represents the harmonic wave of amplitude R.

Now, squaring (3) and (4) and adding, we get

R 2 (cos 2 Ө + sin 2 Ө) = (a + b cosΦ) 2 + (b sinΦ) 2

R 2 .1 = a 2 + b 2 Cos 2 Φ + 2ab cosΦ + b 2 Sin 2 Φ

a 2 + b 2 (Sin 2 Φ+Cos 2 Φ)+2abcosΦ

R = √a 2 + b 2 + 2ab cosΦ

Since, intensity I α  amplitude R 2 .

I  α a 2 + b 2 + 2ab cosΦ

For constructive fringe: 

I should be maximum for which cosΦ = max or +1; Φ = 0, 2π, 4π…

 Φ = 2nπ 

(n = integer such that n = 0,1,2…)

If x is the path difference between the two waves reaching point P (in Fig.2) corresponding to phase difference Φ, then,

              x  = λ/2π Φ = \[\frac{\lambda }{2\pi }(2n\pi )\]

 x = nλ….(p)                  

For destructive interference:

I should be minimum i.e., CosΦ= minimum when Φ = -1 or π, 3π, 5π….

         So,    

Φ = (2n - 1)π 

The path difference will be:

x  =  λ/2π Φ =  λ/2π (2n - 1)π

 x = λ/2(2n - 1)...(q)

Young's Double Slits Formula Derivation

Let S 1 and S 2 be two slits  separated by a distance d, and the center O equidistant from S 1 and S 2 .

Let the slits be illuminated by a monochromatic source S of light of wavelength λ. 

Consider a point P at a distance y from C.

Here, O is the midpoint of  S 1 and S 2 , and

                     D = OP₀ and D >> d

As S 1 S 2   are perpendicular to OP₀ and S 1 A nearly perpendicular to O., we have

                                   ∠ S₁ S₂A = ∠ POP₀ = Ө

The path difference between two waves approaching at P is,

                     Δ x   = S₂P - S₁P = S₂P - PA (Since D>>d)

                              = S₂A = d SinӨ

                              =  d tanӨ = dy/D 

                            Δ x =  dy/D = nλ (where n is an integer)                       

So,  

                             y = nDλ/d…..(a)

At n = 0, y = 0

n = 1, y = Dλ/d

n = 2, y = 2Dλ/d

n  =3, y = Dλ/d,...etc

The centers of the dark fringes will be obtained when 

                      Δ x = dy/D = (2n - 1)λ

                                 y  = (2n -1)Dλ/d….(b)

At n =1, y =  Dλ/2d

n =2, y =  3Dλ/2d,

n  = 3, y = 5Dλ/2d,...etc

Now, to find the fringe width, subtracting equation (b)  from (a), we get

Fringe width,  w  = (2n -1)Dλ/d - nDλ/d = Dλ/d    

YDSE Derivation

If a glass slab of refractive index μ and thickness t is introduced on one of the paths of interfering waves, the optical length of this path will become μ instead of t, increasing by (t-1)μ.

This generates a path difference, given by,

 Δx =  μt - t = (t-1)μ                    

This path difference comes due to the glass slab.

If current position of fringe is y =D/d (Δx ), the new position will be 

                   y’  = D/d(Δx  + (t-1)μ) 

Lateral shift of fringe:

                              y ₀  = y’ - y

                                   = D/d(t-1)μ

As, fringe width, w = D/dλ 

 We get,

   y ₀   = β/λ  (t-1)μ

FAQs on Young's Double Slit Experiment Derivation

1. What is Fringe Width?

Fringe width is the distance between two consecutive dark and bright fringes and is denoted by a symbol, β. 

Fringe width depends on the following factors that are outlined below:

The wavelength of light.

The distance between the slits and the screen or slit separation.

2. What is Fringe Width in YDSE?

The distance between any two consecutive dark or bright fringes and all the fringes are of equal lengths. Fringe width is given by, β = D/dλ.

The angular width, Ө = λd = βD.

3. How is Fringe Width Calculated?

Let’s say the wavelength of the light is 6000 Å. The distance between the two slits is d = 0.8 x 10 -3 m . Distance (D) between slit and screen is 1.2 m. The fringe width will be calculated by the formula: β = Dλ/d  =   1.2 x  6 x 10 -7 /0.8 x 10 -3   ( 1 Å =  10 -10 m)

On calculating, we get

β = 9 x 10 -4 m 

4. What is The Ratio of Fringe Width For Bright And Dark Fringes?

In Young’s double slit experiment, dark and bright fringes are equally spaced. Therefore, the ratio of fringe width for dark to bright fringes is 1.

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Young's Experiment

• Anatomy of a Two-Point Source Interference Pattern
• The Path Difference
• Young's Equation
• Young's Experiment
• Other Applications of Two-Point Source Interference

Today's classroom version of the same experiment is typically performed using a laser beam as the source. Rather than using a note card to split the single beam into two coherent beams, a carbon-coated glass slide with two closely spaced etched slits is used. The slide with its slits is most commonly purchased from a manufacturer who provides a measured value for the slit separation distance - the d value in Young's equation. Light from the laser beam diffracts through the slits and emerges as two separate coherent waves. The interference pattern is then projected onto a screen where reliable measurements can be made of L and y for a given bright spot with order value m . Knowing these four values allows a student to determine the value of the wavelength of the original light source.

To illustrate some typical results from this experiment and the subsequent analysis, consider the sample data provided below for d, y, L and m.

(Note: AN 0 = central antinode and AN 4 = fourth antinode)  

The determination of the wavelength demands that the above values for d, y, L and m be substituted into Young's equation.

Careful inspection of the units of measurement is always advisable. The sample data here reveal that each measured quantity is recorded with a different unit. Before substituting these measured values into the above equation, it is important to give some thought to the treatment of units. One means of resolving the issue of nonuniform units is to simply pick a unit of length and to convert all quantities to that unit. If doing so, one might want to pick a unit that one of the data values already has so that there is one less conversion. A wise choice is to choose the meter as the unit to which all other measured values are converted. Since there are 1000 millimeters in 1 meter, the 0.250 mm is equivalent to 0.000250 meter. And since there are 100 centimeters in 1 meter, the 10.2 cm is equivalent to 0.102 m. Thus, the new values of d, y and L are:

While the conversion of all the data to the same unit is not the only means of treating such measured values, it might be the most advisable - particularly for those students who are less at ease with such conversions.

Now that the issue regarding the units of measurement has been resolved, substitution of the measured values into Young's equation can be performed.

λ  = 6.52 x 10 -7 m

As is evident here, the wavelength of visible light is rather small. For this reason wavelength is often expressed using the unit nanometer, where 1 meter is equivalent to 10 9 nanometers. Multiplying by 10 9 will convert the wavelength from meters to nanometers (abbreviated nm).

We Would Like to Suggest ...

Check Your Understanding

1. The diagram below depicts the results of Young's Experiment. The appropriate measurements are listed on the diagram. Use these measurements to determine the wavelength of light in nanometers. (GIVEN: 1 meter = 10 9 nanometers)

Answer: 657 nm

First, identify known values in terms of their corresponding variable symbol:

L = 10.2 m = 1020 cm y = 22.5 cm m = 10 d = 0.298 mm = 0.0298 cm

(Note: m was chosen as 10 since the y distance corresponds to the distance from the 5th bright band on one side of the central band and the 5th bright band on the other side of the central band.)

Then convert all known values to an identical unit. In this case, cm has been chosen as the unit to use. The converted values are listed in the table above.

Substitute all values into Young's equation and perform calculation of the wavelength. The unit of wavelength is cm.

λ = y • d / ( m • L) λ = ( 22.5 cm ) • ( 0.0298 cm ) / [ ( 10 ) • ( 1020 cm ) ] λ = 6.57 x 10 -5 cm

Finally convert to nanometers using a conversion factor. If there are 10 9 nm in 1 meter, then there must be 10 7 nm in the smaller centimeter.

λ = ( 6.57 x 10 -5 cm ) • ( 10 7 nm / 1 cm ) = 657 nm

2. A student uses a laser and a double-slit apparatus to project a two-point source light interference pattern onto a whiteboard located 5.87 meters away. The distance measured between the central bright band and the fourth bright band is 8.21 cm. The slits are separated by a distance of 0.150 mm. What would be the measured wavelength of light?

Answer: 524 nm

L = 5.87 m = 587 cm y = 8.21 cm m = 4 d = 0.150 mm = 0.0150 cm

λ = y • d / ( m • L) λ = ( 8.21 cm ) • ( 0.0150 cm ) / [ ( 4 ) • ( 587 cm ) ] λ = 5.24 x 10 -5 cm

λ = ( 5.24 x 10 -5 cm ) • ( 10 7 nm / 1 cm ) = 524 nm

3. The analysis of any two-point source interference pattern and a successful determination of wavelength demands an ability to sort through the measured information and equating the values with the symbols in Young's equation. Apply your understanding by interpreting the following statements and identifying the values of y, d, m and L. Finally, perform some conversions of the given information such that all information share the same unit.

This question simply asks to equate the stated information with the variables of Young's equation and to perform conversions such that all information is in the same unit.

(Note that m = 4.5 represents the fifth nodal position or dark band from the central bright band. Also note that the given values have been converted to cm.)

b. An interference pattern is produced when light is incident upon two slits that are 50.0 micrometers apart. The perpendicular distance from the midpoint between the slits to the screen is 7.65 m. The distance between the two third-order antinodes on opposite sides of the pattern is 32.9 cm.

This question simply asks to equate the stated information with the variables of Young's equation and to perform conversions such that all information is in the same unit. 

(Note that m = 6 corresponds to six spacings. There are three spacings between the central antinode and the third antinode. The stated distance is twice as far so the m value must be doubled. Also note that the given values have been converted to cm. There are 10 6 µm in one meter; so there are 10 4 µm in one centimeter.)

c. The fourth nodal line on an interference pattern is 8.4 cm from the first antinodal line when the screen is placed 235 cm from the slits. The slits are separated by 0.25 mm.

 ( Note that the fourth nodal line is assigned the order value of 3.5. Also note that the given values have been converted to cm.)

d. Two sources separated by 0.500 mm produce an interference pattern 525 cm away. The fifth and the second antinodal line on the same side of the pattern are separated by 98 mm.

  ( Note that there are three spacings between the second and the fifth bright bands. Since all spacings are the same distance apart, the distance between the second and the fifth bright bands would be the same as the distance between the central and the third bright bands. Thus, m = 3. Also note that the given values have been converted to cm.)

e. Two slits that are 0.200 mm apart produce an interference pattern on a screen such that the central maximum and the 10th bright band are distanced by an amount equal to one-tenth the distance from the slits to the screen.

 ( Note that there are 10 spacings between the central anti-node and the tenth bright band or tenth anti-node. And observe that they do not state the actual values of L and y; the value of y is expressed in terms of L. )

f. The fifth antinodal line and the second nodal line on the opposite side of an interference pattern are separated by a distance of 32.1 cm when the slits are 6.5 m from the screen. The slits are separated by 25.0 micrometers.

 ( Note that there are five spacings between the central anti-node and the fifth anti-node. And there are 1.5 spacings from the central anti-node in the opposite direction out to the second nodal line. Thus, m = 6.5. Also note that the given values have been converted to cm. There are 10 6 µm in one meter; so there are 10 4 µm in one centimeter.)

g. If two slits 0.100 mm apart are separated from a screen by a distance of 300 mm, then the first-order minimum will be 1 cm from the central maximum.

  ( Note that a the first-order minimum is a point of minimum brightness or a nodal position. The first-order minimum is the first nodal position and is thus the m = 0.5 node. Also note that the given values have been converted to cm. )

h. Consecutive bright bands on an interference pattern are 3.5 cm apart when the slide containing the slits is 10.0 m from the screen. The slit separation distance is 0.050 mm.

   ( Note that the spacing between adjacent bands is given. This distance is equivalent with the distance from the central bright band to the first antinode. Thus, m = 1. Also note that the given values have been converted to cm. )

Interference of Light Waves - Young's Double-slit Experiment

Wave nature of light, parameters and equations, effects of varying certain variables.

Early physicists such as Issac Newton and Christiann Huygens thought tt was a steady stream of particles and had a wave motion. Indeed, such theories were correct but were unproven because light does indeed consists of moving quantas of photons which are packets of particles of pure energy. It was not until 1802, that Thomas Young proved that light does have a wave motion. In his experiment, he hypothesized that parallel beams of light should produce a brighter area where they overlapped if light were a steady stream. Instead, they produced bright bands separated by dark bands, the result of two sets of waves coinciding or canceling each other out. Thus, light is a form of wave.

• Interference is the phenomenon that arises when two or more waves meet at the same place and their amplitudes add in a way that they are either constructive or destructive depending on their relative phases.
• Intereference occurs any time two waves interact, but in order to get a reliable pattern of intereference, coherent or zmonochromatic light sources are needed. This is because coherent light means that the waves emitted are in-phase, whereas monochromatic light emit a single color which means the waves are in-phase for sure. (This is why light bulbs are not used in light waves experiments because they are incoherent.)
• In his experiment, Young allowed sunlight to pass through a small hole, that in turn pass through a pair of closely spaced slits, which then illuminated a screen. Waves diffract at each slit and then interfere in the region between the slits and the screen thus causing a pattern of alternating dark and bright regions on the screen. These regions are called fringes .
• Dark fringes are the result of destructive interference when the waves are out of phase, whereas bright fringes are formed by constructive interence when the waves are in-phase.

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Study: Early dark energy could resolve cosmology’s two biggest puzzles

In the universe’s first billion years, this brief and mysterious force could have produced more bright galaxies than theory predicts..

A new study by MIT physicists proposes that a mysterious force known as early dark energy could solve two of the biggest puzzles in cosmology and fill in some major gaps in our understanding of how the early universe evolved.

One puzzle in question is the “Hubble tension,” which refers to a mismatch in measurements of how fast the universe is expanding. The other involves observations of numerous early, bright galaxies that existed at a time when the early universe should have been much less populated.

Now, the MIT team has found that both puzzles could be resolved if the early universe had one extra, fleeting ingredient: early dark energy. Dark energy is an unknown form of energy that physicists suspect is driving the expansion of the universe today. Early dark energy is a similar, hypothetical phenomenon that may have made only a brief appearance, influencing the expansion of the universe in its first moments before disappearing entirely.

Some physicists have suspected that early dark energy could be the key to solving the Hubble tension, as the mysterious force could accelerate the early expansion of the universe by an amount that would resolve the measurement mismatch.

The MIT researchers have now found that early dark energy could also explain the baffling number of bright galaxies that astronomers have observed in the early universe. In their new study, reported today in the Monthly Notices of the Royal Astronomical Society , the team modeled the formation of galaxies in the universe’s first few hundred million years. When they incorporated a dark energy component only in that earliest sliver of time, they found the number of galaxies that arose from the primordial environment bloomed to fit astronomers’ observations.

“ You have these two looming open-ended puzzles,” says study co-author Rohan Naidu , a postdoc in MIT’s Kavli Institute for Astrophysics and Space Research. “We find that in fact, early dark energy is a very elegant and sparse solution to two of the most pressing problems in cosmology.”

The study’s co-authors include lead author and Kavli postdoc Xuejian (Jacob) Shen, and MIT professor of physics Mark Vogelsberger , along with Michael Boylan-Kolchin at the University of Texas at Austin, and Sandro Tacchella at the University of Cambridge.

Big city lights

Based on standard cosmological and galaxy formation models, the universe should have taken its time spinning up the first galaxies. It would have taken billions of years for primordial gas to coalesce into galaxies as large and bright as the Milky Way.

But in 2023, NASA’s James Webb Space Telescope (JWST) made a startling observation. With an ability to peer farther back in time than any observatory to date, the telescope uncovered a surprising number of bright galaxies as large as the modern Milky Way within the first 500 million years, when the universe was just 3 percent of its current age.

“The bright galaxies that JWST saw would be like seeing a clustering of lights around big cities, whereas theory predicts something like the light around more rural settings like Yellowstone National Park,” Shen says. “And we don’t expect that clustering of light so early on.”

For physicists, the observations imply that there is either something fundamentally wrong with the physics underlying the models or a missing ingredient in the early universe that scientists have not accounted for. The MIT team explored the possibility of the latter, and whether the missing ingredient might be early dark energy.

Physicists have proposed that early dark energy is a sort of antigravitational force that is turned on only at very early times. This force would counteract gravity’s inward pull and accelerate the early expansion of the universe, in a way that would resolve the mismatch in measurements. Early dark energy, therefore, is considered the most likely solution to the Hubble tension.

Galaxy skeleton

The MIT team explored whether early dark energy could also be the key to explaining the unexpected population of large, bright galaxies detected by JWST. In their new study, the physicists considered how early dark energy might affect the early structure of the universe that gave rise to the first galaxies. They focused on the formation of dark matter halos — regions of space where gravity happens to be stronger, and where matter begins to accumulate.

“We believe that dark matter halos are the invisible skeleton of the universe,” Shen explains. “Dark matter structures form first, and then galaxies form within these structures. So, we expect the number of bright galaxies should be proportional to the number of big dark matter halos.”

The team developed an empirical framework for early galaxy formation, which predicts the number, luminosity, and size of galaxies that should form in the early universe, given some measures of “cosmological parameters.” Cosmological parameters are the basic ingredients, or mathematical terms, that describe the evolution of the universe.

Physicists have determined that there are at least six main cosmological parameters, one of which is the Hubble constant — a term that describes the universe’s rate of expansion. Other parameters describe density fluctuations in the primordial soup, immediately after the Big Bang, from which dark matter halos eventually form.

The MIT team reasoned that if early dark energy affects the universe’s early expansion rate, in a way that resolves the Hubble tension, then it could affect the balance of the other cosmological parameters, in a way that might increase the number of bright galaxies that appear at early times. To test their theory, they incorporated a model of early dark energy (the same one that happens to resolve the Hubble tension) into an empirical galaxy formation framework to see how the earliest dark matter structures evolve and give rise to the first galaxies.

“What we show is, the skeletal structure of the early universe is altered in a subtle way where the amplitude of fluctuations goes up, and you get bigger halos, and brighter galaxies that are in place at earlier times, more so than in our more vanilla models,” Naidu says. “It means things were more abundant, and more clustered in the early universe.”

“A priori, I would not have expected the abundance of JWST’s early bright galaxies to have anything to do with early dark energy, but their observation that EDE pushes cosmological parameters in a direction that boosts the early-galaxy abundance is interesting,” says Marc Kamionkowski, professor of theoretical physics at Johns Hopkins University, who was not involved with the study. “I think more work will need to be done to establish a link between early galaxies and EDE, but regardless of how things turn out, it’s a clever — and hopefully ultimately fruitful — thing to try.”

“ We demonstrated the potential of early dark energy as a unified solution to the two major issues faced by cosmology. This might be an evidence for its existence if the observational findings of JWST get further consolidated,” Vogelsberger concludes. “In the future, we can incorporate this into large cosmological simulations to see what detailed predictions we get.”

This research was supported, in part, by NASA and the National Science Foundation.

• Paper: “Early Galaxies and Early Dark Energy: A Unified Solution to the Hubble Tension and Puzzles of Massive Bright Galaxies revealed by JWST”

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