IB Physics Quiz: Understand Wave Phenomena

20 questions · exam conditions

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Understand Wave PhenomenaQuestion 1 of 20

Light with a wavelength of 600 nm in a vacuum (n=1.00) enters a block of glass with a refractive index of 1.50. It then immediately enters a second material with a refractive index of 1.20. What is the wavelength of the light in the second material?

400 nm

500 nm

720 nm

900 nm

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IB Physics Quiz

IB Physics Quiz: Understand Wave Phenomena

Practice Understand Wave Phenomena in IB Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Understand Wave Phenomena, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

400 nm
500 nm (correct answer)
720 nm
900 nm

Explanation: The frequency of light remains constant when it changes medium. The wavelength in a medium with refractive index

n

is given by

\lambda_{medium} = \frac{\lambda_{vacuum}}{n}

. The properties of the first block of glass are irrelevant to the final wavelength in the second material, as long as the reference vacuum wavelength is known. Therefore, the wavelength in the second material is

\lambda_2 = \frac{600 \text{ nm}}{1.20} = 500 \text{ nm}

Question 2

Significant diffraction occurs when a wave passes through an aperture. By what factor must the frequency of a wave be changed for it to diffract to the same extent through an aperture that is twice as wide? Assume the wave speed is constant.

It must be quadrupled.
It must be doubled.
It must be halved. (correct answer)
It must be quartered.

Explanation: The extent of diffraction depends on the ratio of the wavelength to the aperture size,

\lambda/b

. For the diffraction to be the same, this ratio must remain constant. If the aperture width

b

is doubled (

b' = 2b

), the wavelength

\lambda

must also be doubled (

\lambda' = 2\lambda

) to keep the ratio

\lambda'/b'

the same. Since wave speed

v

is constant and

v = f\lambda

, frequency is inversely proportional to wavelength (

f = v/\lambda

). To double the wavelength, the frequency must be halved.

Question 3

In a double-slit experiment, red light ( $\lambda_R = 700$ nm) produces a first-order bright fringe at a certain position on a screen. If the red light is replaced by blue light ( $\lambda_B = 400$ nm), which bright fringe of the blue light will be formed closest to the original position of the red light's first-order fringe?

First-order
Second-order (correct answer)
Third-order
Fourth-order

Explanation: The position of the

n

-th order bright fringe is given by

x = \frac{n\lambda D}{d}

. The position of the first-order red fringe is

x_R = \frac{1 \cdot \lambda_R D}{d} = \frac{700 D}{d}

(in arbitrary units). We want to find an integer

n_B

such that the position of the

n_B

-th blue fringe,

x_B = \frac{n_B \lambda_B D}{d} = \frac{n_B \cdot 400 D}{d}

, is closest to

x_R

. We set

x_B \approx x_R

, so

n_B \cdot 400 \approx 700

. This gives

n_B \approx 700/400 = 1.75

. The closest integer to 1.75 is 2. Therefore, the second-order blue fringe will be closest.

Question 4

[HL Only]

Monochromatic light is incident on a diffraction grating with slit separation $d$ . The third-order maximum is observed at an angle of 30° to the central maximum. What is the ratio of the wavelength to the slit separation, $\lambda/d$ ?

1/6 (correct answer)
1/3
$\sqrt{3}/6$
3/2

Explanation: The condition for a maximum in a diffraction grating is given by the equation

d \sin\theta = n\lambda

, where

n

is the order of the maximum. We are given

n=3

and

\theta = 30°

. Substituting these values gives

d \sin(30°) = 3\lambda

. Since

\sin(30°) = 0.5 = 1/2

, the equation becomes

d (1/2) = 3\lambda

. We are asked for the ratio

\lambda/d

. Rearranging the equation gives

\frac{\lambda}{d} = \frac{1/2}{3} = \frac{1}{6}

Question 5

A sound wave travels from warm air into colder air. The speed of sound is higher in warm air than in cold air. As the wave crosses the boundary, what happens to its frequency and wavelength?

Frequency is constant; Wavelength decreases (correct answer)
Frequency is constant; Wavelength increases
Frequency decreases; Wavelength is constant
Frequency increases; Wavelength decreases

Explanation: The frequency of a wave is determined by its source and remains constant as it propagates through different media. The wave relationship is

v = f\lambda

. Since the sound wave travels from warm air (higher speed

v

) to cold air (lower speed

v

), its speed decreases. Because

f

is constant, if

v

decreases, the wavelength

\lambda

must also decrease to maintain the equality.

Question 6

A light ray travels in a material with refractive index $n_1 = 2.0$ . It is incident on a boundary with a second material with refractive index $n_2 = 1.4$ . What is the approximate critical angle for total internal reflection?

35°
70°
46°
44° (correct answer)

Explanation: Total internal reflection can occur because the light is traveling from a higher refractive index medium to a lower one (

n_1 > n_2

). The critical angle

\theta_c

is defined by the equation

n_1 \sin\theta_c = n_2 \sin(90°)

. This simplifies to

\sin\theta_c = n_2 / n_1

. Plugging in the values:

\sin\theta_c = 1.4 / 2.0 = 0.70

. To find the angle, we take the inverse sine:

\theta_c = \arcsin(0.70) \approx 44.4°

. The closest answer is 44°.

Question 7

Two coherent point sources S1 and S2 emit waves of the same wavelength $\lambda$ . A point P is located such that the distance S1P is 12.5 $\lambda$ and the distance S2P is 10.0 $\lambda$ . What is the nature of the interference at point P?

Constructive, producing a maximum.
Destructive, producing a minimum. (correct answer)
Partially destructive, between a maximum and minimum.
It is impossible to determine without knowing the phase of the sources.

Explanation: Interference depends on the path difference between the two waves. The path difference is

\Delta x = |S1P - S2P| = |12.5\lambda - 10.0\lambda| = 2.5\lambda

. Destructive interference occurs when the path difference is an odd integer multiple of half a wavelength, i.e.,

(m + \frac{1}{2})\lambda

. Here,

2.5\lambda = (2 + \frac{1}{2})\lambda

, which fits the condition for destructive interference with

m=2

Question 8

[HL Only]

In a single-slit diffraction experiment, the angle subtended by the first minimum is $\theta$ . If the experiment is repeated with light of double the wavelength and a slit of four times the width, what is the new angle $\theta'$ subtended by the first minimum?

$\theta/4$
$2\theta$
$\theta$
$\theta/2$ (correct answer)

Explanation: For a single slit, the condition for the first minimum is

b \sin\theta = \lambda

. For small angles, this is approximated as

b\theta \approx \lambda

, so

\theta \approx \lambda/b

. The initial angle is

\theta \approx \lambda/b

. The new parameters are

\lambda' = 2\lambda

and

b' = 4b

. The new angle is

\theta' \approx \lambda'/b' = (2\lambda)/(4b) = \frac{1}{2}(\lambda/b) = \theta/2

Question 9

A single slit of width $b$ is illuminated by light of wavelength $\lambda$ . A diffraction pattern is formed on a distant screen. If the width of the slit is decreased, what happens to the width of the central maximum and the intensity of the central maximum?

Width increases; Intensity increases
Width increases; Intensity decreases (correct answer)
Width decreases; Intensity increases
Width decreases; Intensity decreases

Explanation: The angular position of the first minimum in a single-slit diffraction pattern is given by

\sin\theta = \lambda/b

. The width of the central maximum is proportional to

2\theta

. If the slit width

b

is decreased,

\sin\theta

increases, so the angle

\theta

increases. This means the central maximum becomes wider. However, by making the slit narrower, less light energy passes through per unit time, so the overall intensity of the pattern, including the central maximum, decreases.

Question 10

In a Young's double-slit experiment, the separation between adjacent bright fringes is $s$ . The entire apparatus is then submerged in a transparent liquid with a refractive index $n > 1$ . What is the new fringe separation?

$n^2 s$
$n s$
$s$
$s/n$ (correct answer)

Explanation: The fringe separation is given by

s = \frac{\lambda D}{d}

. When the apparatus is submerged in a liquid of refractive index

n

, the wavelength of the light changes to

\lambda' = \frac{\lambda}{n}

. The distance to the screen

D

and the slit separation

d

remain unchanged. The new fringe separation

s'

will be

s' = \frac{\lambda' D}{d} = \frac{(\lambda/n) D}{d} = \frac{1}{n} \left(\frac{\lambda D}{d}\right) = \frac{s}{n}

Question 11

Two wave pulses travel towards each other on a string. One pulse has a displacement that varies from 0 to +5 cm. The other pulse has a displacement that varies from 0 to -3 cm. What is the maximum possible magnitude of the displacement of the string during the interaction?

2 cm
5 cm (correct answer)
8 cm
15 cm

Explanation: The principle of superposition states that the resultant displacement is the algebraic sum of the individual displacements. The question asks for the maximum possible magnitude. The interaction will produce various displacements as the pulses pass through each other. The maximum positive displacement occurs when the peak of the +5 cm pulse aligns with a zero-displacement part of the -3 cm pulse, giving +5 cm. The maximum negative displacement occurs when the trough of the -3 cm pulse aligns with a zero-displacement part of the +5 cm pulse, giving -3 cm. The magnitude of the displacement is maximized at 5 cm.

Question 12

[HL Only]

A diffraction grating has N slits per unit length. For a fixed wavelength, what is the effect of increasing N on the angular separation of the principal maxima and on their sharpness?

Angular separation increases; Sharpness increases (correct answer)
Angular separation increases; Sharpness decreases
Angular separation decreases; Sharpness increases
Angular separation decreases; Sharpness decreases

Explanation: The slit separation

d

is the reciprocal of the number of slits per unit length,

d = 1/N

. The grating equation is

d \sin\theta = n\lambda

, or

\sin\theta = n\lambda N

. If N increases,

\sin\theta

increases for a given order

n

, meaning the angular separation between orders increases. The sharpness of the principal maxima is proportional to the total number of slits illuminated. Increasing N means more slits are illuminated over the same beam width, which causes more effective destructive interference between the maxima, making them sharper (narrower).

Question 13

Light is refracted at the boundary between two media, with an angle of incidence of 45° and an angle of refraction of 30°. What is the ratio of the speed of light in the first medium to the speed of light in the second medium (v₁/v₂)?

$\frac{\sin 30°}{\sin 45°}$
$\frac{\sin 45°}{\sin 30°}$ (correct answer)
$\frac{\cos 30°}{\cos 45°}$
$\frac{\cos 45°}{\cos 30°}$

Explanation: According to Snell's Law,

n_1 \sin \theta_1 = n_2 \sin \theta_2

. The refractive index

n

is defined as

n = c/v

, where

c

is the speed of light in a vacuum and

v

is the speed in the medium. Substituting this into Snell's law gives

\frac{c}{v_1} \sin \theta_1 = \frac{c}{v_2} \sin \theta_2

. Rearranging for the ratio

v_1/v_2

gives

\frac{v_1}{v_2} = \frac{\sin \theta_1}{\sin \theta_2}

. Given

\theta_1 = 45°

and

\theta_2 = 30°

, the ratio is

\frac{\sin 45°}{\sin 30°}

Question 14

Coherent light of wavelength $\lambda$ is incident on two slits separated by a distance $d$ . A screen is placed a distance $D$ from the slits. The path difference between the waves arriving at the second-order dark fringe from the central maximum is

$1.5 \lambda$ (correct answer)
$2.0 \lambda$
$2.5 \lambda$
$3.0 \lambda$

Explanation: Destructive interference, which creates a dark fringe, occurs when the path difference is an odd integer multiple of half a wavelength, i.e., path difference =

(m + \frac{1}{2})\lambda

, where

m = 0, 1, 2, ...

. The first-order dark fringe corresponds to

m=0

(path difference

0.5\lambda

). The second-order dark fringe corresponds to

m=1

, so the path difference is

(1 + \frac{1}{2})\lambda = 1.5\lambda

Question 15

A wave pulse travels along a string and encounters a boundary where the string density changes. The incident pulse has amplitude A and the reflected pulse has amplitude 0.6A. What fraction of the incident wave's energy is transmitted through the boundary?

0.36
0.64 (correct answer)
0.40
0.80

Explanation: Wave energy is proportional to the square of amplitude. The reflected amplitude is 0.6A, so the reflected energy fraction is

(0.6)^2 = 0.36

. By conservation of energy, the transmitted energy fraction is

1 - 0.36 = 0.64

. Choice A incorrectly uses only the reflected energy. Choice C uses the amplitude ratio directly. Choice D incorrectly assumes amplitude and energy have the same relationship.

Question 16

A circular water wave spreads outward from a point source. At a distance of 2.0 m from the source, the wave amplitude is 5.0 cm. Assuming no energy loss, what will be the amplitude when the wave reaches a distance of 8.0 m from the source?

1.25 cm (correct answer)
2.5 cm
5.0 cm
10.0 cm

Explanation: For a circular wave spreading in two dimensions, energy is conserved but distributed over an increasing circumference. The intensity (energy per unit area) decreases as

\frac{1}{r^2}

, and since intensity is proportional to amplitude squared, amplitude decreases as

\frac{1}{r}

. Therefore:

\frac{A_2}{A_1} = \frac{r_1}{r_2} = \frac{2.0}{8.0} = 0.25

. So

A_2 = 5.0 \times 0.25 = 1.25

cm. Choice B uses

\frac{1}{\sqrt{r}}

relationship. Choice C assumes no amplitude change. Choice D incorrectly uses inverse relationship.

Question 17

A standing wave is formed on a string fixed at both ends. The string has length 1.2 m and the third harmonic has a frequency of 450 Hz. If the string tension is doubled while keeping the linear mass density constant, what will be the frequency of the second harmonic?

424 Hz (correct answer)
600 Hz
636 Hz
900 Hz

Explanation: For a string fixed at both ends,

f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}

. Initially,

f_3 = 450

Hz, so the fundamental frequency is

f_1 = \frac{450}{3} = 150

Hz. When tension doubles, the wave speed increases by

\sqrt{2}

, so all frequencies increase by

\sqrt{2}

. The new fundamental frequency is

150\sqrt{2} \approx 212

Hz. The second harmonic is

2 \times 212 = 424

Hz. Choice B assumes linear relationship with tension. Choice C uses

\sqrt{2} \times 450

. Choice D doubles the original third harmonic frequency.

Question 18

A wave traveling in medium 1 with speed $v_1 = 300$ m/s strikes a boundary with medium 2 at an angle of 30° to the normal. The refracted wave travels at 45° to the normal in medium 2. If a wave pulse in medium 1 has a frequency of 150 Hz, what will be the wavelength of the transmitted pulse in medium 2?

1.41 m
2.00 m
2.83 m (correct answer)
4.24 m

Explanation: Using Snell's law:

\frac{\sin 30°}{v_1} = \frac{\sin 45°}{v_2}

, so

\frac{0.5}{300} = \frac{0.707}{v_2}

. Solving:

v_2 = \frac{0.707 \times 300}{0.5} = 424

m/s. The frequency remains constant at 150 Hz during transmission. The wavelength in medium 2 is

\lambda_2 = \frac{v_2}{f} = \frac{424}{150} = 2.83

m. Choice A uses the wrong speed calculation. Choice B assumes equal speeds. Choice D incorrectly multiplies instead of dividing.

Question 19

For two light sources to produce a stable interference pattern, they must be coherent. Which statement gives the essential conditions for coherence?

The sources have the same amplitude and a phase difference of zero.
The sources have the same frequency and a constant phase difference. (correct answer)
The sources emit waves that are polarized in the same plane.
The sources are point sources and emit light of the same intensity.

Explanation: Coherence has two requirements: 1) the waves must have the same frequency (be monochromatic), and 2) the phase difference between the waves at any given point must be constant over time. The phase difference does not need to be zero. Amplitude and intensity affect the visibility of the fringes, and polarization affects whether interference can occur, but they are not the defining conditions for coherence itself.

Question 20

Which of the following phenomena provides the strongest evidence for the wave nature of light rather than its particle nature?

Reflection from a mirror
Refraction through a lens
Diffraction through a narrow slit (correct answer)
The photoelectric effect

Explanation: Diffraction is the spreading of waves as they pass through an aperture or around an obstacle. This phenomenon is characteristic of waves and cannot be explained by a simple particle model of light (where particles would travel in straight lines). Reflection and refraction can be explained by both wave and particle models (Newton's corpuscular theory explained them). The photoelectric effect provides strong evidence for the particle nature of light (photons).