# 27.3 Young’s double slit experiment  (Page 5/10)

 Page 5 / 10

## Test prep for ap courses

Superposition of which of the following light waves may produce interference fringes? Select two answers.

Wave 1 = A 1 sin(2 ωt )

Wave 2 = A 2 sin(4 ωt )

Wave 3 = A 3 sin(2 ωt + θ )

Wave 4 = A 4 sin(4 ωt + θ ).

1. Wave 1 and Wave 2
2. Wave 2 and Wave 4
3. Wave 3 and Wave 1
4. Wave 4 and Wave 3

(b) and (c)

In a double slit experiment with monochromatic light, the separation between the slits is 2 mm. If the screen is moved by 100 mm toward the slits, the distance between the central bright line and the second bright line changes by 32 μm. Calculate the wavelength of the light used for the experiment.

In a double slit experiment, a student measures the maximum and minimum intensities when two waves with equal amplitudes are used. The student then doubles the amplitudes of the two waves and performs the measurements again. Which of the following will remain unchanged?

1. The intensity of the bright fringe
2. The intensity of the dark fringe
3. The difference in the intensities of consecutive bright and dark fringes
4. None of the above

(b)

Draw a figure to show the resultant wave produced when two coherent waves (with equal amplitudes x ) interact in phase. What is the amplitude of the resultant wave? If the phase difference between the coherent waves is changed to 60º, what will be new amplitude?

What will be the amplitude of the central fringe if the amplitudes of the two waves in a double slit experiment are a and 3 a ?

1. 2 a
2. 4 a
3. 8 a 2
4. 16 a 2

(b)

If the ratio of amplitudes of the two waves in a double slit experiment is 3:4, calculate the ratio of minimum intensity (dark fringe) to maximum intensity (bright fringe).

## 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 $d\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\mathrm{m\lambda }\phantom{\rule{0.25em}{0ex}}\text{(for}\phantom{\rule{0.25em}{0ex}}m=0,\phantom{\rule{0.25em}{0ex}}1,\phantom{\rule{0.25em}{0ex}}-1,\phantom{\rule{0.25em}{0ex}}2,\phantom{\rule{0.25em}{0ex}}-2,\phantom{\rule{0.25em}{0ex}}\dots \right)$ , where $d$ is the distance between the slits, $\theta$ is the angle relative to the incident direction, and $m$ is the order of the interference.
• There is destructive interference when $d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\left(m+\frac{1}{2}\right)\lambda \phantom{\rule{0.25em}{0ex}}\text{(for}\phantom{\rule{0.5em}{0ex}}m=0,\phantom{\rule{0.25em}{0ex}}1,\phantom{\rule{0.25em}{0ex}}-1,\phantom{\rule{0.25em}{0ex}}2,\phantom{\rule{0.25em}{0ex}}-2,\phantom{\rule{0.25em}{0ex}}\dots \right)$ .

## 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.

[link] 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

At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?

$0\text{.}\text{516º}$

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 $\text{30}\text{.}0º$ ?

$1\text{.}\text{22}×{\text{10}}^{-6}\phantom{\rule{0.25em}{0ex}}\text{m}$

Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of $\text{45}\text{.}0º$ .

Calculate the wavelength of light that has its third minimum at an angle of $\text{30}\text{.}0º$ when falling on double slits separated by $3\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μm}$ . Explicitly, show how you follow the steps in Problem-Solving Strategies for Wave Optics .

600 nm

What is the wavelength of light falling on double slits separated by $2\text{.}\text{00}\phantom{\rule{0.25em}{0ex}}\text{μm}$ if the third-order maximum is at an angle of $\text{60}\text{.}0º$ ?

At what angle is the fourth-order maximum for the situation in [link] ?

$2\text{.}\text{06º}$

What is the highest-order maximum for 400-nm light falling on double slits separated by $\text{25}\text{.}0\phantom{\rule{0.25em}{0ex}}\text{μm}$ ?

Find the largest wavelength of light falling on double slits separated by $1\text{.}\text{20}\phantom{\rule{0.25em}{0ex}}\text{μm}$ for which there is a first-order maximum. Is this in the visible part of the spectrum?

1200 nm (not visible)

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) 760 nm

(b) 1520 nm

(a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of $\text{10}\text{.}0º$ , 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?

[link] shows a double slit located a distance $x$ from a screen, with the distance from the center of the screen given by $y$ . When the distance $d$ between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, for small angles (where $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta \approx \theta$ , with $\theta$ in radians), the distance between fringes is given by $\text{Δ}y=\mathrm{x\lambda }/d$ .

For small angles $\text{sin}\phantom{\rule{0.25em}{0ex}}\theta -\text{tan}\phantom{\rule{0.25em}{0ex}}\theta \approx \theta \phantom{\rule{0.25em}{0ex}}\left(\text{in radians}\right)$ .

For two adjacent fringes we have,

$d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{m}}=\mathrm{m\lambda }$

and

$d\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{m}+1}=\left(m+1\right)\lambda$

Subtracting these equations gives

$\begin{array}{}d\left(\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{m}+1}-\text{sin}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{m}}\right)=\left[\left(m+1\right)-m\right]\lambda \\ d\left({\theta }_{\text{m}+1}-{\theta }_{\text{m}}\right)=\lambda \\ \text{tan}\phantom{\rule{0.25em}{0ex}}{\theta }_{\text{m}}=\frac{{y}_{\text{m}}}{x}\approx {\theta }_{\text{m}}⇒d\left(\frac{{y}_{\text{m}+1}}{x}-\frac{{y}_{\text{m}}}{x}\right)=\lambda \\ d\frac{\Delta y}{x}=\lambda ⇒\Delta y=\frac{\mathrm{x\lambda }}{d}\end{array}$

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 [link] .

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 [link] ).

450 nm

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