Diffraction

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AP Physics 2 › Diffraction

Questions 1 - 10
1

A student compares two ripple-tank setups with identical barriers and slit widths. Setup 1 uses waves with longer wavelength than setup 2. Both have the same amplitude. Which setup shows greater diffraction after the slit?

Setup 1, because the wavelength is larger relative to the slit width

Setup 1, because the amplitude is the same so diffraction increases

Setup 2, because the wave speed is higher for shorter wavelengths

Setup 2, because the wavelength is smaller relative to the slit width

Explanation

This question tests understanding of diffraction. Diffraction is the spreading of waves through openings, with the amount of spreading depending on the ratio of wavelength to opening size. Setup 1, with longer wavelength waves passing through the same slit width, will show greater diffraction because the wavelength is larger relative to the slit width. The diffraction angle is proportional to λ/a, where λ is wavelength and a is slit width, so larger wavelengths produce more spreading. Choice B represents the misconception that smaller wavelengths diffract more, but actually smaller wavelengths relative to the opening result in less spreading. Remember that diffraction is strongest when the opening is comparable to the wavelength.

2

Ocean waves pass through a narrow gap between two piers. The gap width is about equal to the wavelength $\lambda$, and the waves spread into the region beyond. Compared to a much wider gap, the spreading is greater mainly because the gap is

transmitting waves of larger amplitude, increasing diffraction

smaller relative to $\lambda$, increasing diffraction

increasing the wave frequency beyond the gap, increasing diffraction

making the waves travel faster beyond the gap, increasing diffraction

Explanation

This question tests understanding of diffraction. Diffraction is the spreading of waves when they pass through openings, with maximum spreading occurring when the opening size is comparable to or smaller than the wavelength. The narrow gap between piers causes significant spreading because its width is about equal to λ, creating strong diffraction. The waves spread out in a semicircular pattern beyond the gap, reaching areas that would be in shadow if waves traveled only in straight lines. Choice B incorrectly suggests that wave amplitude affects diffraction, but diffraction depends solely on the geometric relationship between wavelength and opening size, not on wave intensity. Remember that diffraction is strongest when the opening is comparable to the wavelength.

3

A ripple tank sends plane water waves toward a barrier with a single slit. The slit width is about the same as the wavelength $\lambda$, and the wave speed is unchanged. Which situation shows the greatest diffraction after the slit?

Use a narrower slit while keeping $\lambda$ the same

Increase the wave amplitude while keeping slit width and $\lambda$ the same

Use a wider slit while keeping $\lambda$ the same

Increase the wave speed while keeping slit width and $\lambda$ the same

Explanation

This question tests understanding of diffraction. Diffraction is the spreading of waves when they pass through an opening or around an obstacle, and the amount of spreading depends on the ratio of wavelength to opening size. When the slit width is comparable to the wavelength, significant diffraction occurs, with maximum spreading when the slit becomes narrower relative to the wavelength. The wave spreads out more with a narrower slit because the opening constrains the wavefront more severely, causing greater bending. Choice B incorrectly suggests amplitude affects diffraction, which is a common misconception - diffraction depends only on the wavelength-to-opening ratio, not on wave intensity. To maximize diffraction, make the opening size comparable to or smaller than the wavelength.

4

Microwaves of wavelength $\lambda$ pass through an opening into a lab. When the opening is made comparable to $\lambda$, the transmitted wavefront becomes nearly semicircular. Which condition produces the greatest diffraction?

Higher wave speed at the same opening width

Opening width $\gg \lambda$

Opening width $\approx \lambda$

Higher transmitted amplitude at the same opening width

Explanation

This question tests understanding of diffraction. Diffraction is the spreading of waves when they pass through openings, with maximum spreading occurring when the opening size is comparable to the wavelength. When the opening width approximately equals λ, the transmitted waves spread into a nearly semicircular pattern, indicating strong diffraction. This happens because the opening acts almost like a point source, allowing waves to spread in all forward directions. Choice B incorrectly suggests that much larger openings cause more diffraction, but when the opening is much larger than the wavelength, waves pass through with minimal bending. The fundamental principle is that diffraction is strongest when the opening is comparable to the wavelength.

5

A single slit of width $a$ produces a diffraction pattern on a distant screen. The wavelength $\lambda$ is fixed, and the central maximum is observed. Which change makes the pattern spread out more?

Increase the wave speed while keeping $a$ and $\lambda$ the same

Increase the amplitude while keeping $a$ and $\lambda$ the same

Decrease $a$ while keeping $\lambda$ the same

Increase $a$ while keeping $\lambda$ the same

Explanation

This question tests understanding of diffraction. Diffraction through a single slit produces a pattern whose spread depends on the ratio λ/a, where λ is wavelength and a is slit width. The angular position of the first minimum is given by sin θ = λ/a, so decreasing the slit width a while keeping λ constant increases the angle θ, making the pattern spread out more. This occurs because a narrower slit constrains the wavefront more, causing greater bending as waves emerge. Choice B incorrectly assumes amplitude affects diffraction spreading, but diffraction patterns depend only on the wavelength-to-slit ratio, not on wave intensity. To increase diffraction spreading, decrease the slit width relative to the wavelength.

6

Light of wavelength $500\ \text{nm}$ passes through a slit of width $a$ and forms a diffraction pattern on a far screen. The light source intensity is doubled while $\lambda$ and $a$ stay the same. Compared to before, the angular width of the central maximum is

smaller, because brighter light travels faster

larger, because increased intensity increases diffraction

unchanged, because it depends on $\lambda$ relative to $a$

larger, because amplitude determines spreading angle

Explanation

This question tests understanding of diffraction. Diffraction patterns depend on the ratio of wavelength to slit width (λ/a), which determines the angular width of features like the central maximum. When only the light intensity is changed while keeping wavelength and slit width constant, the λ/a ratio remains unchanged, so the angular width of the central maximum stays the same. The pattern becomes brighter but maintains the same angular dimensions. Choice A incorrectly suggests intensity affects diffraction angle, but intensity only affects brightness, not the geometric spreading of the pattern. The key principle is that diffraction geometry depends only on the wavelength-to-opening ratio, not on wave intensity.

7

Water waves with wavelength $\lambda$ pass through a single slit of fixed width $w$. The student repeats the trial using waves of wavelength $2\lambda$ while keeping the slit width and water depth the same. In both trials, the wave spreads after passing through the slit. Compared to the $\lambda$ trial, how does the spreading change for the $2\lambda$ trial?

It is unchanged, because the slit width is the same

It is greater, because the ratio $\lambda/w$ is larger

It is smaller, because the wave speed is unchanged

It is greater, because the wave amplitude is larger for longer wavelengths

Explanation

This question tests understanding of diffraction. Diffraction occurs when waves encounter an opening or obstacle, with the amount of spreading determined by the ratio λ/w, where λ is wavelength and w is the opening width. When wavelength increases while the slit width remains constant, the ratio λ/w increases, resulting in greater diffraction and more spreading of the waves. In this case, doubling the wavelength from λ to 2λ while keeping w constant doubles the ratio λ/w, causing increased spreading. Choice C incorrectly assumes that diffraction depends only on slit width, ignoring the crucial role of wavelength - this represents a misconception about what controls diffraction. The transferable strategy is that diffraction increases when wavelength increases relative to the opening size.

8

Ocean waves of wavelength about $20\ \text{m}$ approach a breakwater with a gap. In case 1 the gap is $5\ \text{m}$ wide; in case 2 the gap is $40\ \text{m}$ wide. Wave height and speed are the same. Which case produces the greatest bending into the sheltered region?

Case 2, because the waves have higher intensity through the gap

Case 1, because the gap is much smaller than the wavelength

Case 1, because the wave speed decreases in narrower openings

Case 2, because a wider gap increases diffraction

Explanation

This question tests understanding of diffraction. Diffraction is the bending and spreading of waves as they pass through openings, with maximum spreading when the opening is comparable to or smaller than the wavelength. The ocean waves have wavelength 20 m, while Case 1 has a 5 m gap (much smaller than wavelength) and Case 2 has a 40 m gap (larger than wavelength). Case 1 produces greater diffraction because the gap (5 m) is much smaller than the wavelength (20 m), causing waves to spread dramatically into the sheltered region. Choice A incorrectly claims wider gaps increase diffraction, but actually narrower gaps relative to wavelength increase diffraction. The principle is that diffraction is strongest when the opening is comparable to or smaller than the wavelength.

9

Two identical slits each of width $a$ are used separately with different light. Setup R uses red light ($\lambda_R$) and setup V uses violet light ($\lambda_V$), with $\lambda_R>\lambda_V$ and equal brightness. Which setup produces the greater diffraction (wider central maximum) on a far screen?

Setup R, because greater intensity causes more spreading

Setup R, because larger wavelength relative to $a$ increases diffraction

Setup V, because higher frequency increases diffraction

Setup V, because violet light travels faster in air

Explanation

This question tests understanding of diffraction. Diffraction is the spreading of light as it passes through slits, with the angular width of the central maximum proportional to λ/a (wavelength divided by slit width). Since red light has a longer wavelength than violet light (λR > λV) and both pass through identical slits of width a, red light has a larger λ/a ratio and produces a wider central maximum. Choice A incorrectly claims higher frequency increases diffraction, but higher frequency means shorter wavelength, which actually decreases diffraction for a fixed slit width. The key principle is that diffraction increases as wavelength increases relative to the opening size.

10

A microwave of wavelength $\lambda$ passes through a rectangular opening of width $w$. In Setup A, $w=\lambda/3$; in Setup B, $w=4\lambda$. After the opening, the wavefront spreads. Which setup produces the greatest diffraction?

Setup B, because the transmitted amplitude is larger

Setup A, because the wave speed decreases in the opening

Setup A, because $w$ is much smaller than $\lambda$

Setup B, because larger openings increase bending of waves

Explanation

This question tests understanding of diffraction. Diffraction is the bending and spreading of waves as they pass through an opening, with the effect being most pronounced when the opening width w is small compared to the wavelength λ. In Setup A, w = λ/3 means the opening is much smaller than the wavelength, causing significant diffraction and spreading. In Setup B, w = 4λ means the opening is much larger than the wavelength, resulting in minimal diffraction with waves continuing mostly straight. Choice A represents a common misconception that larger openings cause more bending, when the opposite is true - smaller openings relative to wavelength cause greater diffraction. The key principle is that diffraction is strongest when the opening is comparable to or smaller than the wavelength.

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