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Young's Double-Slit Experiment ( AQA A Level Physics )

Revision note.

Double Slit Interference

The interference of two coherent wave sources
A single wave source passing through a double slit
The laser light source is placed behind the single slit
So the light is diffracted, producing two light sources at slits A and B
The light from the double slits is then diffracted, producing a diffraction pattern made up of bright and dark fringes on a screen

3-3-3--laser-light-double-slit-experiment-diagram

The typical arrangement of Young's double-slit experiment

Diffraction Pattern

Constructive interference between light rays forms bright strips, also called fringes , interference fringes or maxima , on the screen
Destructive interference forms dark strips, also called dark fringes or minima , on the screen

$youngs-double-slit-diffraction-pattern$

Young's double slit experiment and the resulting diffraction pattern

Each bright fringe is identical and has the same width and intensity
The fringes are all separated by dark narrow bands of destructive interference

$creation-of-diffraction-pattern-aqa-al-physics$

The constructive and destructive interference of laser light through a double slit creates bright and dark strips called fringes on a screen placed far away

Interference Pattern

The Young's double slit interference pattern shows the regions of constructive and destructive interference:
Each bright fringe is a peak of equal maximum intensity
Each dark fringe is a a trough or minimum of zero intensity
The maxima are formed by the constructive interference of light
The minima are formed by the destructive interference of light

The interference pattern of Young's double-slit diffraction of light

When two waves interfere, the resultant wave depends on the path difference between the two waves
This extra distance is the path difference

Path difference equations, downloadable AS & A Level Physics revision notes

The path difference between two waves is determined by the number of wavelengths that cover their difference in length

For constructive interference (or maxima), the difference in wavelengths will be an integer number of whole wavelengths
For destructive interference (or minima) it will be an integer number of whole wavelengths plus a half wavelength
There is usually more than one produced
n is the order of the maxima or minima; which represents the position of the maxima away from the central maximum
n = 0 is the central maximum
n = 1 represents the first maximum on either side of the central, n = 2 the next one along....

Worked example

WE - Two source interference question image, downloadable AS & A Level Physics revision notes

Determine which orders of maxima are detected at M as the wavelength is increased from 3.5 cm to 12.5 cm.

Worked example - two source interference (2), downloadable AS & A Level Physics revision notes

The path difference is more specifically how much longer, or shorter, one path is than the other. In other words, the difference in the distances. Make sure not to confuse this with the distance between the two paths.

Fringe Spacing Equation

The spacing between the bright or dark fringes in the diffraction pattern formed on the screen can be calculated using the double slit equation:

Double slit interference equation with w, s and D represented on a diagram

D is much bigger than any other dimension, normally several metres long
s is the separation between the two slits and is often the smallest dimension, normally in mm
w is the distance between the fringes on the screen, often in cm. This can be obtained by measuring the distance between the centre of each consecutive bright spot.
The wavelength , λ of the incident light increases
The distance , s between the screen and the slits increases
The separation , w between the slits decreases

WE - Double slit equation question image, downloadable AS & A Level Physics revision notes

Calculate the separation of the two slits.

Fringe Spacing Worked Example, downloadable AS & A Level Physics revision notes

Since w , s and D are all distances, it's easy to mix up which they refer to. Labelling the double-slit diagram with each of these quantities can help ensure you don't use the wrong variable for a quantity.

Interference Patterns

It is different to that produced by a single slit or a diffraction grating

$diffraction-of-white-light-double-slit$

The interference pattern produced when white light is diffracted through a double slit

Each maximum is of roughly equal width
There are two dark narrow destructive interference fringes on either side
All other maxima are composed of a spectrum
The shortest wavelength (violet / blue) would appear nearest to the central maximum because it is diffracted the least
The longest wavelength (red) would appear furthest from the central maximum because it is diffracted the most
As the maxima move further away from the central maximum, the wavelengths of blue observed decrease and the wavelengths of red observed increase

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Author: Katie M

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.

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).

A beam of light strikes a wall through which a pair of vertical slits is cut. On the other side of the wall, another wall shows a pattern of equally spaced vertical lines of light that are of the same height as the slit.

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 a shows three sine waves with the same wavelength arranged one above the other. The peaks and troughs of each wave are aligned with those of the other waves. The top two waves are labeled wave one and wave two and the bottom wave is labeled resultant. The amplitude of waves one and two are labeled x and the amplitude of the resultant wave is labeled two x. Figure b shows a similar situation, except that the peaks of wave two now align with the troughs of wave one. The resultant wave is now a straight horizontal line on the x axis; that is, the line y equals zero.

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.

The figure contains three parts. The first part is a drawing that shows parallel wavefronts approaching a wall from the left. Crests are shown as continuous lines, and troughs are shown as dotted lines. Two light rays pass through small slits in the wall and emerge in a fan-like pattern from two slits. These lines fan out to the right until they hit the right-hand wall. The points where these fan lines hit the right-hand wall are alternately labeled min and max. The min points correspond to lines that connect the overlapping crests and troughs, and the max points correspond to the lines that connect the overlapping crests. The second drawing is a view from above of a pool of water with semicircular wavefronts emanating from two points on the left side of the pool that are arranged one above the other. These semicircular waves overlap with each other and form a pattern much like the pattern formed by the arcs in the first image. The third drawing shows a vertical dotted line, with some dots appearing brighter than other dots. The brightness pattern is symmetric about the midpoint of this line. The dots near the midpoint are the brightest. As you move from the midpoint up, or down, the dots become progressively dimmer until there seems to be a dot missing. If you progress still farther from the midpoint, the dots appear again and get brighter, but are much less bright than the central dots. If you progress still farther from the midpoint, the dots get dimmer again and then disappear again, which is where the dotted line stops.

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.

Both parts of the figure show a schematic of a double slit experiment. Two waves, each of which is emitted from a different slit, propagate from the slits to the screen. In the first schematic, when the waves meet on the screen, one of the waves is at a maximum whereas the other is at a minimum. This schematic is labeled dark (destructive interference). In the second schematic, when the waves meet on the screen, both waves are at a minimum.. This schematic is labeled bright (constructive interference).

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?

The figure is a schematic of a double slit experiment, with the scale of the slits enlarged to show the detail. The two slits are on the left, and the screen is on the right. The slits are represented by a thick vertical line with two gaps cut through it a distance d apart. Two rays, one from each slit, angle up and to the right at an angle theta above the horizontal. At the screen, these rays are shown to converge at a common point. The ray from the upper slit is labeled l sub one, and the ray from the lower slit is labeled l sub two. At the slits, a right triangle is drawn, with the thick line between the slits forming the hypotenuse. The hypotenuse is labeled d, which is the distance between the slits. A short piece of the ray from the lower slit is labeled delta l and forms the short side of the right triangle. The long side of the right triangle is formed by a line segment that goes downward and to the right from the upper slit to the lower ray. This line segment is perpendicular to the lower ray, and the angle it makes with the hypotenuse is labeled theta. Beneath this triangle is the formula delta l equals d sine theta.

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.

The figure consists of two parts arranged side-by-side. The diagram on the left side shows a double slit arrangement along with a graph of the resultant intensity pattern on a distant screen. The graph is oriented vertically, so that the intensity peaks grow out and to the left from the screen. The maximum intensity peak is at the center of the screen, and some less intense peaks appear on both sides of the center. These peaks become progressively dimmer upon moving away from the center, and are symmetric with respect to the central peak. The distance from the central maximum to the first dimmer peak is labeled y sub one, and the distance from the central maximum to the second dimmer peak is labeled y sub two. The illustration on the right side shows thick bright horizontal bars on a dark background. Each horizontal bar is aligned with one of the intensity peaks from the first figure.

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.

The figure shows a photo of a horizontal line of equally spaced red dots of light on a black background. The central dot is the brightest and the dots on either side of center are dimmer. The dot intensity decreases to almost zero after moving six dots to the left or right of center. If you continue to move away from the center, the dot brightness increases slightly, although it does not reach the brightness of the central dot. After moving another six dots, or twelve dots in all, to the left or right of center, there is another nearly invisible dot. If you move even farther from the center, the dot intensity again increases, but it does not reach the level of the previous local maximum. At eighteen dots from the center, there is another nearly invisible dot.

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?

The figure shows a schematic of a double slit experiment. A double slit is at the left and a screen is at the right. The slits are separated by a distance d. From the midpoint between the slits, a horizontal line labeled x extends to the screen. From the same point, a line angled upward at an angle theta above the horizontal also extends to the screen. The distance between where the horizontal line hits the screen and where the angled line hits the screen is marked y, and the distance between adjacent fringes is given by delta y, which equals x times lambda over d.

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.

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

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

$\boldsymbol{d \;\textbf{sin} \;\theta = m +}$

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

$\boldsymbol{d \;\textbf{sin} \;\theta = m \theta , \;\textbf{for} \; m = 0, \;1, \; -1, \; 2, \; -2, \; \dots}.$

Example 1: Finding a Wavelength from an Interference Pattern

$\boldsymbol{10.95 ^{\circ}}$

Substituting known values yields

$$\begin{array}{r @{{}={}}l} \boldsymbol{\lambda} & \boldsymbol{\frac{(0.0100 \;\textbf{mm})(\textbf{sin} 10.95^{\circ})}{3}} \\[1em] & \boldsymbol{6.33 \times 10^{-4} \;\textbf{mm} = 633 \;\textbf{nm}}. \end{array}$

Example 2: Calculating Highest Order Possible

Strategy and Concept

$\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda \; (\textbf{for} \; m = 0, \; 1, \; -1, \; 2, \; -2, \; \dots)}$

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.

$\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda \;(\textbf{for} \; m = 0, \; 1, \; -1, \;2, \; -2, \dots)}$

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.

$\boldsymbol{30.0 ^{\circ}}$

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

$\boldsymbol{25.0 \;\mu \textbf{m}}$

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?

$\boldsymbol{10.0^{\circ}}$

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 ).

$\boldsymbol{0.516 ^{\circ}}$

9: 1200 nm (not visible)

11: (a) 760 nm

(b) 1520 nm

$\boldsymbol{\textbf{sin} \;\theta - \;\textbf{tan} \;\theta \approx \theta}$

For two adjacent fringes we have,

$\boldsymbol{d \;\textbf{sin} \;\theta _{\textbf{m}} = m \lambda}$

Subtracting these equations gives

$\begin{array}{r @{{}={}}l} \boldsymbol{d (\textbf{sin} \; \theta _{\textbf{m} + 1} - \textbf{sin} \; \theta _{\textbf{m}})} & \boldsymbol{[(m + 1) - m] \lambda} \\[1em] \boldsymbol{d(\theta _{{\textbf{m}} + 1} - \theta _{\textbf{m}})} & \boldsymbol{\lambda} \end{array}$

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In Young's double-slit experiment, the angular width of a fringe formed on a distant screen is 1 o . The wavelength of light used is 6000 ˚ A . What is the spacing between the slits? 344 m m 0.0344 m m 0.1344 m m 0.034 m m

Angular fringe width: β θ = λ d = 6000 × 10 − 10 d ∴ d = 6 × 10 − 7 1 × π 180 = 0.344 × 10 − 4 m = 0.0344 m.

Interference: Young’s Double Slit Experiment
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Optics of Class 12

Interference: young’s double slit experiment .

It was carried out in 1802 by the English scientist Thomas Young to prove the wave nature of light.

Two slits S 1 and S 2 are made in an opaque screen, parallel and very close to each other. These two are illuminated by another narrow slits S which in turn lit by a bright source. Light wave spread out from S and fall on both S 1 and S 2 which behave like coherent sources. Note that the coherent sources are derived from the same source. In this way, any phase change in S occurs in both S 1 and S 2 . The phase difference (φ 1 - φ 2 ) between S 1 and S 2 is unaffected and remain constant.

Light now spreads out from both S 1 and S 2 falls on a screen. It is essential that the waves from the two sources overlap on the same part of the screen. If one slit is covered up, the other produces a wide smoothly illuminated patch on the screen. But when both slits are open, the patch is seen to be crossed by dark and bright bands called interference fringes.

Condition for bright fringes or maxima

φ = 2nπ or path difference , p = nλ where n = 0, 1, 2, ……

Interference: Young’s Double Slit Experiment

Condition for dark fringes or minima

φ = (2n – 1)πorpath difference, p = (n - 1/2)λ

where n = 1, 2, 3, …………

The relation between phase difference (φ) and path difference (p) is given by

φ = 2π/λ p (15.32)

How to find the position of the nth maxima or minima on the screen?

Let P be the position of the nth maxima on the screen. The two waves arriving at P follow the path S 1 P and S 2 P, thus the path difference between the two waves is

p = S 1 P – S 1 P = d sin θ

From experimental conditions, we know that D >> d, therefore, the angle θ is small,

thus sinθ ≈ tanθ = Yn/D

∴p = d sinθ = d tan θ = d (Yn/D) ⇒ y n = pD/d

For nth maxima

p = nλ

∴y n = nλ D/d where n = 0, 1, 2, ……..(15.33)

For nth minima

p = (n - 1/2)λ

∴y n = (n - 1/2)λD/d where n = 1, 2, 3, ……(15.34)

Note that the nth minima comes before the nth maxima.

Fringe Width

It is defined as the distance between two successive maxima or minima.

Example 15.23

In a YDSE , if D = 2 m; d = 6 mm and λ = 6000 Å, then

(a)find the fringe width

(b)find the position of the 3 rd maxima

(c)find the position of the 2 nd minima

Solution

(a)Using equation (15.33),

ω = λD/d = (6000 x 10 -10 )(2)/6 x 10 -3 = 0.2 mm

(b)Position of 3 rd maxima is obtained by putting n = 3 in the equation (15.33),

y 3 = 3 λD = 3ω = 3(0.2) = 0.6 mm

(c)Position of 2 nd minima is obtained by putting n = 2 in the equation (15.34),

Example 15.24

What is the effect on the interference fringes in a YDSE due to each of the following operations.

(a)the screen is moved away from the plane of the slits

(b)the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength

(c)the separation between the two slits is increased

(d)the monochromatic source is replaced by source of white light

(e)the whole experiment is carried out in a medium of refractive index μ

(a)Angular separation (= λ/d ) of the fringes remains constant. But the linear separation or fringe width increases in proportion to the distance (D) from the screen.

(b)As λ decreases, fringe width (ω ∝ λ) decreases

(c)As d increases, fringe width (ω ∝ 1/d) decreases

(d)The interference pattern due to different component colours of white light overlap (in-coherently). The central bright fringes of different colours are at the same position. Therefore, the central fringe is white. Since blue colour has the lower λ, the fringe closest on either side of the central white fringe is blue; the farthest is red.

Thus, fringe width decreases by μ.

Example 15.25

Light from a source consists of two wavelength λ 1 = 6500Å and λ 2 = 5200Å. If D = 2m and d = 6.5 mm find the minimum value of y (≠ 0) where the maxima of both the wavelengths coincide.

Let n 1 th maxima of λ 1 coincides with n 2 th maxima of λ 2 .

Then yn1 = yn2

Thus, fourth maxima of λ 1 coincides with fifth maxima of λ 2 .

The minimum value of y (≠ 0) is given by

Optical Path

It is defined as distance travelled by light in vacuum in the same time in which it travels a given path length in a medium. If light travels a path length d in a medium at speed c, the time taken by it will be (d/c). So optical path length−

L = c o × [d/c] = μd(because μ = c0/c )(15.36)

Since for all media μ > 1, optical path length is always greater than geometrical path length. When two light waves arrive at a point by travelling different distances in different media, the phase difference between the two is related by their optical path difference instead of simply path difference.

Phase Difference = 2π/ λ (optical path difference) = 2π/ λ L

Fringe Shift

When a transparent film of thickness t and refractive index μ is introduced in front of one of the slits, the fringe pattern shifts in the direction where the film is placed.

How much is the fringe shift?

Consider the YDSE arrangement shown in the figure.

The optical path difference is given by

p = [(S2P – t) + μt] – S1P

orp = (S2P – S1P) + t(μ - 1)

SinceS2P – S1P = d sin θ

∴p = d sin θ + t (μ - 1)

As sin θ ≈ tan θ =

∴p =

In the absence of film, the position of the nth maxima is given by equation (15.34)

y n = n λD/d

Therefore, the fringe shift is given by

FS = y n - y′ n = (μ - 1) tD/d (15.37)

Note that the shift is in the direction where the film is introduced.

Example 15.26

Ιn a YDSE, λ = 6000Å, D = 2 m, d = 6 mm. When a film of refractive index 1.5 is introduced in front of the lower slit, the third maxima shifts to the origin.

(a)Find the thickness of the film

(b)Find the positions of the fourth maxima

(a)Since 3 rd minima shifts to the origin, therefore, the fringe shift is given by

FS = y 3 = 3 λD/d

From equation (15.37), we know

FS = (μ - 1) tD/d

∴(μ - 1) tD/d = 3 λD/d

or t = 3 λ/μ - 1

Here λ = 0.6 × 10 -6 m; μ = 1.5

∴t = 3(0.6 x 10 -6 )/1.5 -1 = 3.6 μm

(b)There are two positions of 4th maxima: one above and the other below the origin.

y4 = 1ω = λD/d = 0.2 mm

y′4 = -7ω = -7λD/d = 1.4 mm

Intensity Distribution

When two coherent light waves of intensity I 1 and I 2 with a constant phase difference φ superimpose, then the resultant intensity is given by

I = I 1 + I 2 + 2 √I 1 I 2 cosφ

In YDSE, usually the intensities I 1 and I 2 are equal

i.e.I 1 = I 2 = I o

Then I = 2I o ( 1+ cos φ)

orI = 4I o cos 2 (φ/2)(15.38)

For maxima:φ = 2nπ ⇒I max = 4I o

For minima:φ = (2n – 1)π⇒I min = 0

Example 15.27

The intensity of the light coming from one of the slits in YDSE is double the intensity from the other slit. Find the ratio of the maximum intensity to the minimum intensity in the interference fringe pattern observed.

SinceI 1 = 2I o and I 2 = I o , therefore

Example 15.28

In a YDSE, λ = 60 nm; D = 2m; λ = 6 mm. Find the positions of a point lying between third maxima and third minima where the intensity is three-fourth of the maximum intensity on the screen.

Using equation (15.32)

I = 4I o cos 2 φ/2

here I = 3/4 (4I o ) = 3I o

∴cos φ/2 = √3/2

Thus,φ/2 = nπ ± π/6

or φ = 2nπ ± π/3

For the point lying between third minima and third maxima,

or y 3 = 17/6 λD/d

Putting λ = 0.6 × 10 -6 m; D = 2m; d = 6 mm

Geometrical Optics
Reflection from a Plane-surface
Reflection from Curved Surfaces
Refraction Through Plane Surfaces
Refraction on Curved Surfaces
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Table Of Contents

The discovery of light's wave-particle duality

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.)

in young's double slit experiment the angular width

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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.

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Chapter 27 Wave Optics

215 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.

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]\boldsymbol{\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.

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]\boldsymbol{(1/2) \;\lambda}[/latex], [latex]\boldsymbol{(3/2) \;\lambda}[/latex], [latex]\boldsymbol{(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]\boldsymbol{\lambda}[/latex], [latex]\boldsymbol{2 \lambda}[/latex], [latex]\boldsymbol{3 \lambda}[/latex], etc.), then constructive interference occurs.

Take-Home Experiment: Using Fingers as Slits

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]\boldsymbol{\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]\boldsymbol{d \;\textbf{sin} \;\theta}[/latex], where [latex]\boldsymbol{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]\boldsymbol{\lambda}[/latex] is the wavelength of the light, [latex]\boldsymbol{d}[/latex] is the distance between slits, and [latex]\boldsymbol{\theta}[/latex] is the angle from the original direction of the beam as discussed above. We call [latex]\boldsymbol{m}[/latex] the order of the interference. For example, [latex]\boldsymbol{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]\boldsymbol{\lambda}[/latex] and [latex]\boldsymbol{m}[/latex], the smaller [latex]\boldsymbol{d}[/latex] is, the larger [latex]\boldsymbol{\theta}[/latex] must be, since [latex]\boldsymbol{\textbf{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]\boldsymbol{d}[/latex] apart) is small. Small [latex]\boldsymbol{d}[/latex] gives large [latex]\boldsymbol{\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]\boldsymbol{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]\boldsymbol{m = 3}[/latex]. We are given [latex]\boldsymbol{d = 0.0100 \;\textbf{mm}}[/latex] and [latex]\boldsymbol{\theta = 10.95^{\circ}}[/latex]. The wavelength can thus be found using the equation [latex]\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda}[/latex] for constructive interference.

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

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

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

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

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]\boldsymbol{d \;\textbf{sin} \;\theta = m \lambda \;(\textbf{for} \; m = 0, \; 1, \; -1, \;2, \; -2, \dots)}[/latex], where [latex]\boldsymbol{d}[/latex] is the distance between the slits, [latex]\boldsymbol{\theta}[/latex] is the angle relative to the incident direction, and [latex]\boldsymbol{m}[/latex] is the order of the interference.
There is destructive interference when [latex]\boldsymbol{d \;\textbf{sin} \;\theta = (m+ \frac{1}{2}) \lambda}[/latex] (for [latex]\boldsymbol{m = 0, \; 1, \; -1, \; 2, \; -2, \; \dots}[/latex]).

Conceptual Questions

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

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]\boldsymbol{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]\boldsymbol{45.0 ^{\circ}}[/latex].

5: Calculate the wavelength of light that has its third minimum at an angle of [latex]\boldsymbol{30.0 ^{\circ}}[/latex] when falling on double slits separated by [latex]\boldsymbol{3.00 \;\mu \textbf{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]\boldsymbol{2.00 \;\mu \textbf{m}}[/latex] if the third-order maximum is at an angle of [latex]\boldsymbol{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]\boldsymbol{25.0 \;\mu \textbf{m}}[/latex]?

9: Find the largest wavelength of light falling on double slits separated by [latex]\boldsymbol{1.20 \;\mu \textbf{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]\boldsymbol{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]\boldsymbol{x}[/latex] from a screen, with the distance from the center of the screen given by [latex]\boldsymbol{y}[/latex]. When the distance [latex]\boldsymbol{d}[/latex] between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, for small angles (where [latex]\boldsymbol{\textbf{sin} \;\theta \approx \theta}[/latex], with [latex]\boldsymbol{\theta}[/latex] in radians), the distance between fringes is given by [latex]\boldsymbol{\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]\boldsymbol{0.516 ^{\circ}}[/latex]

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

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

9: 1200 nm (not visible)

11: (a) 760 nm

(b) 1520 nm

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

For two adjacent fringes we have,

Subtracting these equations gives

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In a Young's double slit experiment, an angular width of the fringe is 0.35° on a screen placed at 2 m

In a Young's double slit experiment, an angular width of the fringe is 0.35° on a screen placed at 2 m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is 1/α The value of a is. ____

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In YDSE experiment, the angular width of a fringe formed on a distant screen is0.50. If light of wavelength 7000 A is used in the experiment, then distancebetween slits isOptions0.08 mm0.08 cm0.01 mm0.02 cm

IMAGES

Double-Slit Experiment: Explanation, Diagram, and Equation
Young’s Double Slits Experiment Derivation
YOUNG’S DOUBLE SLIT EXPERIMENT PART 01
Young's Double Slit Experiment
Diagram of Young's double slit experiment indicating the intensity of
Young's Double Slit Experiment

COMMENTS

Young's Double Slit Experiment
Fringe Width. The distance between two adjacent bright (or dark) fringes is called the fringe width. β = λD/d. 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.
Young's Double-Slit Experiment
Double Slit Interference. Young's double-slit experiment produces a diffraction and an interference pattern using either: The interference of two coherent wave sources. A single wave source passing through a double slit. In this typical set-up for Young's double slit experiment: The laser light source is placed behind the single slit.
27.3 Young's Double Slit Experiment
Figure 27.10 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.
Young's Double Slit Experiment
Section Summary. Young's double slit experiment gave definitive proof of the wave character of light. An interference pattern is obtained by the superposition of light from two slits. There is constructive interference when d sin θ = mλ ( for m = 0, 1, −1, 2, −2, . . . ), where d is the distance between the slits, θ is the angle ...
27.3 Young's Double Slit Experiment
Section Summary. Young's double slit experiment gave definitive proof of the wave character of light. An interference pattern is obtained by the superposition of light from two slits. There is constructive interference when d sin θ = mλ (for m = 0, 1, −1, 2, −2,…) d sin θ = m λ ( for m = 0, 1, − 1, 2, − 2, …), where d d is the ...
Young's Double Slit Experiment
This physics video tutorial provides a basic introduction into young's double slit experiment. It explains how to calculate the distance between the slits g...
27.3 Young's Double Slit Experiment
We illustrate the double slit experiment with monochromatic (single ) 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.
4.3 Double-Slit Diffraction
When we studied interference in Young's double-slit experiment, we ignored the diffraction effect in each slit. ... The plot shows the expected result for a slit width a = 2 ... From Equation 4.1, the angular position of the first diffraction minimum is ...
Fringe width in Young's double slit experiment
Fringe width is the distance between two consecutive bright spots (maximas, where constructive interference take place) or two consecutive dark spots (minima...
Young's interference experiment
Unlike the modern double-slit experiment, Young's experiment reflects sunlight (using a steering mirror) through a small hole, and splits the thin beam in half using a paper card. [6] [8] [9] He also mentions the possibility of passing light through two slits in his description of the experiment: Modern illustration of the double-slit experiment
In Young's double-slit experiment, the angular width of a ...
In Young's double-slit experiment the spacing between the slits is 'd and wavelength of light used is 6000 ˚ A. If the angular width of a fringe formed on a distant screen is 1 0 , then value ′ d ′ is :
Double-slit experiment
In modern physics, the double-slit experiment demonstrates that light and matter can satisfy the seemingly incongruous classical definitions for both waves and particles. This ambiguity is considered evidence for the fundamentally probabilistic nature of quantum mechanics.This type of experiment was first performed by Thomas Young in 1801, as a demonstration of the wave behavior of visible ...
Interference: Young's Double Slit Experiment
Putting λ = 0.6 × 10 -6 m; D = 2m; d = 6 mm. y 3 =. Question of Class 12-Interference: Young's Double Slit Experiment : It was carried out in 1802 by the English scientist Thomas Young to prove the wave nature of light. Two slits S1 and S2 are made in an opaque screen, parallel and very close to each other. These two are illuminated by ...
Light
Learn about Thomas Young's double-slit experiment that challenged Isaac Newton's theory of light. (more) See all videos for this article. ... 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 ...
27.3 Young's Double Slit Experiment
Section Summary. Young's double slit experiment gave definitive proof of the wave character of light. An interference pattern is obtained by the superposition of light from two slits. There is constructive interference when d sin θ = mλ (for m = 0, 1, −1, 2, −2,…) d sin θ = m λ ( for m = 0, 1, − 1, 2, − 2, …), where d d is the ...
In a double slit experiment the angular width of a fringe is found to
In Young's double slit experiment, the two slits are illuminated by light of wavelength 5890 ∘ A and the angular fringe width is 0.2 ∘. If the whole apparatus is immersed in water, then the new angular fringe width will be (refractive index of water = 4 3).
In a Young's double slit experiment, an angular width of the fringe is
In a Young's double slit experiment, an angular width of the fringe is 0.35° on a screen placed at 2 m away for particular wavelength of 450 nm. The angular width of the fringe, when whole system is immersed in a medium of refractive index 7/5, is 1/α The value of a is.
In YDSE experiment, the angular width of a fringe formed on a distant
In Young's double-slit experiment the spacing between the slits is 'd and wavelength of light used is 6000 ˚ A. If the angular width of a fringe formed on a distant screen is 1 0, then value ′ d ′ is :
In Young's double-slit experiment the angular width of a fringe formed
In Young's double-slit experiment the angular width of a fringe formed on a distant screen is `1^(@)`. The wavelength of light used is `6000 Å`. What is the ...

Young's Double-Slit Experiment ( AQA A Level Physics )

Double Slit Interference

Diffraction Pattern

Interference Pattern

Worked example

Fringe Spacing Equation

Interference Patterns

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Author: Katie M

Wave Optics

Take-Home Experiment: Using Fingers as Slits

Example 1. Finding a Wavelength from an Interference Pattern

Example 2. Calculating Highest Order Possible

Strategy and Concept

Section Summary

Conceptual Questions

Problems & Exercises

Selected Solutions to Problems & Exercises

27.3 Young’s Double Slit Experiment

Take-Home Experiment: Using Fingers as Slits

Example 1: Finding a Wavelength from an Interference Pattern

Example 2: Calculating Highest Order Possible

Section Summary

Conceptual Questions

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In Young's double-slit experiment, the angular width of a fringe formed on a distant screen is 1 o . The wavelength of light used is 6000 ˚ A . What is the spacing between the slits? 344 m m 0.0344 m m 0.1344 m m 0.034 m m

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Fringe Shift

Intensity Distribution

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Young’s double-slit experiment

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215 27.3 Young’s Double Slit Experiment

Take-Home Experiment: Using Fingers as Slits

Example 1: Finding a Wavelength from an Interference Pattern

Example 2: Calculating Highest Order Possible

Section Summary

Conceptual Questions

Share This Book

In a Young's double slit experiment, an angular width of the fringe is 0.35° on a screen placed at 2 m

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In YDSE experiment, the angular width of a fringe formed on a distant screen is0.50. If light of wavelength 7000 A is used in the experiment, then distancebetween slits isOptions0.08 mm0.08 cm0.01 mm0.02 cm

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