IB Physics Quiz: Apply Wave Phenomena

20 questions · exam conditions

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Apply Wave PhenomenaQuestion 1 of 20

For a stable interference pattern to be formed by the light from two sources, the sources must be coherent. This means the waves they emit must have

the same amplitude and be in phase.

the same frequency and a constant phase difference.

the same amplitude and the same frequency.

a phase difference of zero and travel at the same speed.

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IB Physics Quiz

IB Physics Quiz: Apply Wave Phenomena

Practice Apply 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 Apply 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

For a stable interference pattern to be formed by the light from two sources, the sources must be coherent. This means the waves they emit must have

the same amplitude and be in phase.
the same frequency and a constant phase difference. (correct answer)
the same amplitude and the same frequency.
a phase difference of zero and travel at the same speed.

Explanation: Coherence is the property that requires the sources to have a constant phase difference and the same frequency. While having the same amplitude provides for maximum contrast in the interference pattern, it is not a requirement for coherence itself. Being exactly in phase (zero phase difference) is a specific case of coherence, but not the general definition.

Question 2

Light passes from air (n ≈ 1.00) into a rectangular block of glass (n = 1.50). The light then emerges from the parallel opposite side of the block into water (n = 1.33). If the angle of incidence in the air is 60°, what is the final angle of refraction in the water?

35.3°
40.6° (correct answer)
48.8°
77.9°

Explanation: For parallel interfaces, the intermediate medium does not affect the relationship between the initial and final angles. Applying Snell's Law between the initial medium (air) and the final medium (water):

n_{air} \sin \theta_{air} = n_{water} \sin \theta_{water}

. This gives

1.00 \times \sin(60°) = 1.33 \times \sin \theta_{water}

. Solving for

\theta_{water}

gives

\arcsin(\frac{\sin(60°)}{1.33}) \approx 40.6°

Question 3

In a double-slit experiment, the distance between the slits is halved, and the distance from the slits to the screen is doubled. If the original fringe separation was $s$ , what is the new fringe separation?

$s/4$
$s$
$2s$
$4s$ (correct answer)

Explanation: The fringe separation is given by the formula

s = \lambda D / d

, where

D

is the distance to the screen and

d

is the slit separation. The new distance to the screen is

D' = 2D

, and the new slit separation is

d' = d/2

. The new fringe separation

s'

s' = \lambda D' / d' = \lambda (2D) / (d/2) = 4 (\lambda D / d) = 4s

Question 4

Light of wavelength 600 nm is used in a double-slit experiment, producing fringes with a separation of 3.0 mm. If the light source is replaced with one that produces fringes with a separation of 2.0 mm, without changing the experimental setup, what is the wavelength of the new light source?

400 nm (correct answer)
450 nm
800 nm
900 nm

Explanation: Fringe separation

s

is directly proportional to wavelength

\lambda

(from

s = \lambda D/d

). Therefore, the ratio of separations equals the ratio of wavelengths:

s_1 / s_2 = \lambda_1 / \lambda_2

. Plugging in the values:

3.0 / 2.0 = 600 \text{ nm} / \lambda_2

. Solving for

\lambda_2

gives

\lambda_2 = 600 \text{ nm} \times (2.0 / 3.0) = 400 \text{ nm}

Question 5

A small object is at the bottom of a pool of water of depth D. When viewed from directly above, its apparent depth is d. What is the ratio of the speed of light in water to the speed of light in air?

d/D (correct answer)
D/d
$\sqrt{d/D}$
$\sqrt{D/d}$

Explanation: The refractive index of water is given by the ratio of real depth to apparent depth,

n_{water} = D/d

. The refractive index is also the ratio of the speed of light in vacuum/air (c) to the speed of light in the medium (v),

n_{water} = c/v_{water}

. The question asks for

v_{water}/c

, which is

1/n_{water}

. Therefore, the ratio is

1 / (D/d) = d/D

Question 6

In a Young's double-slit experiment, a point on the screen is equidistant from both slits. At this point, there is a bright fringe. A second point on the screen corresponds to the third dark fringe from the center. What is the path difference between the waves arriving at this second point?

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

Explanation: Dark fringes (destructive interference) occur when the path difference is

(m + 1/2)\lambda

, where

m = 0, 1, 2, ...

. The central fringe is at

m=0

for a bright fringe. The first dark fringe corresponds to

m=0

, with a path difference of

0.5\lambda

. The second dark fringe corresponds to

m=1

, path difference

1.5\lambda

. The third dark fringe corresponds to

m=2

, with a path difference of

(2 + 1/2)\lambda = 2.5\lambda

Question 7

Two wave pulses travel toward each other on a rope. One pulse has a positive displacement of amplitude 3.0 cm. The other has a negative displacement of amplitude 1.5 cm. What is the resultant displacement of the rope at the instant the pulses completely overlap?

4.5 cm
3.0 cm
1.5 cm (correct answer)
0 cm

Explanation: According to the principle of superposition, the resultant displacement at any point is the algebraic sum of the individual displacements. At the point of complete overlap, the resultant displacement is

(+3.0 \text{ cm}) + (-1.5 \text{ cm}) = +1.5 \text{ cm}

Question 8

When unpolarized light passes through two polarizing filters, the intensity of the transmitted light is at a minimum. What is the angle between the transmission axes of the two filters?

0°
45°
90° (correct answer)
180°

Explanation: The first filter polarizes the unpolarized light, reducing its intensity by half. The second filter is an analyzer. According to Malus's Law, the intensity transmitted through the second filter is

I = I_0 \cos^2\theta

, where

\theta

is the angle between the axes of the two filters. The intensity is at a minimum (zero) when

\cos^2\theta = 0

, which occurs when

\theta = 90°

270°

. This is known as crossed polarizers.

Question 9

Two coherent sources S1 and S2 emit monochromatic waves of wavelength $\lambda$ . They produce an interference pattern. At a point P, the path difference (S2P - S1P) is $3.5\lambda$ . What is observed at point P?

A maximum intensity, as it is a point of constructive interference.
A minimum intensity, as it is a point of destructive interference. (correct answer)
An intensity that is halfway between maximum and minimum.
An intensity that depends on the phase of the sources.

Explanation: Constructive interference (maximum intensity) occurs when the path difference is an integer multiple of the wavelength (

n\lambda

). Destructive interference (minimum intensity) occurs when the path difference is a half-integer multiple of the wavelength (

(n+1/2)\lambda

). Since the path difference is

3.5\lambda

, which is a half-integer multiple, destructive interference occurs, resulting in a minimum intensity.

Question 10

A monochromatic light wave travels from a medium with refractive index $n_1$ to a medium with refractive index $n_2$ , where $n_1 > n_2$ . What are the changes in the wave's speed and wavelength?

Speed increases and wavelength increases. (correct answer)
Speed decreases and wavelength decreases.
Speed increases and wavelength decreases.
Speed decreases and wavelength increases.

Explanation: The frequency of the wave remains constant as it crosses the boundary. The speed of light in a medium is given by

v = c/n

. Since

n_1 > n_2

, the speed

v

increases as the light enters the second medium. Because

v = f\lambda

and

f

is constant, if

v

increases, the wavelength

\lambda

must also increase.

Question 11

Light travels from water (n=1.33) into a layer of oil (n=1.47), which rests on top of glass (n=1.52). At which interface(s) is total internal reflection impossible, regardless of the angle of incidence?

The water-oil interface only.
The oil-glass interface only.
Both the water-oil and oil-glass interfaces. (correct answer)
Neither interface; it is possible at both.

Explanation: Total internal reflection can only occur when light travels from a medium of higher refractive index to a medium of lower refractive index. At the water-oil interface, light travels from n=1.33 to n=1.47 (lower to higher), so TIR is impossible. At the oil-glass interface, light travels from n=1.47 to n=1.52 (lower to higher), so TIR is also impossible.

Question 12

A double-slit experiment is performed in air, producing an interference pattern with fringe spacing $s_{air}$ . The entire apparatus, including the source, slits, and screen, is then submerged in a transparent liquid with refractive index $n > 1$ . What is the new fringe spacing, $s_{liquid}$ ?

$s_{liquid} = n \times s_{air}$
$s_{liquid} = s_{air} / n$ (correct answer)
$s_{liquid} = n^2 \times s_{air}$
$s_{liquid} = s_{air} / n^2$

Explanation: Fringe spacing is given by

s = \lambda D / d

. When the experiment is submerged, the wavelength of the light changes. The wavelength in the liquid,

\lambda_{liquid}

, is related to the wavelength in air,

\lambda_{air}

, by

\lambda_{liquid} = \lambda_{air} / n

. Since D and d are unchanged, the new fringe spacing will be

s_{liquid} = \lambda_{liquid} D / d = (\lambda_{air}/n) D/d = s_{air} / n

Question 13

A ray of light is incident from air onto the surface of a glass block. Which statement is always true regarding the angles of incidence ( $\theta_i$ ), reflection ( $\theta_r$ ), and refraction ( $\theta_t$ )?

$\theta_i = \theta_r$ and $\theta_i = \theta_t$
$\theta_i = \theta_r$ and $\theta_i > \theta_t$ (correct answer)
$\theta_i > \theta_r$ and $\theta_i = \theta_t$
$\theta_i = \theta_r$ and $\theta_i < \theta_t$

Explanation: The law of reflection states that the angle of incidence equals the angle of reflection (

\theta_i = \theta_r

). According to Snell's law, when light enters a more optically dense medium (like glass from air), it bends towards the normal. This means the angle of refraction will be smaller than the angle of incidence (

\theta_i > \theta_t

Question 14

Sound waves of frequency 500 Hz and light waves from a red laser are directed towards an open doorway of width 1.0 m. The speed of sound is 340 m s⁻¹. Diffraction of the sound is easily observed but diffraction of the light is not. What is the reason for this difference?

The intensity of the light is much greater than the intensity of the sound.
Sound waves are longitudinal whereas light waves are transverse.
The speed of sound is far less than the speed of light.
The wavelength of the sound is comparable to the doorway's width, while the light's wavelength is much smaller. (correct answer)

Explanation: Significant diffraction occurs when the wavelength of a wave is on the order of the size of the aperture or obstacle. The wavelength of the sound is

\lambda = v/f = 340/500 = 0.68

m, which is comparable to the 1.0 m doorway. The wavelength of red light is approximately 700 nm (

7 \times 10^{-7}

m), which is vastly smaller than the doorway. Therefore, sound diffracts noticeably while light does not.

Question 15

A beam of white light travels within a glass block and strikes the glass-air boundary at an angle of incidence that is gradually increased. Due to dispersion, the refractive index of the glass is slightly different for different colors. Which color of light is the first to undergo total internal reflection?

Red, because it has the longest wavelength.
Violet, because it has the highest frequency.
Red, because it has the lowest refractive index in glass.
Violet, because it has the highest refractive index in glass. (correct answer)

Explanation: The critical angle is given by

\sin \theta_c = 1/n

. Due to dispersion in glass, the refractive index (n) is largest for violet light (shortest wavelength) and smallest for red light (longest wavelength). A larger refractive index results in a smaller critical angle. Since violet light has the smallest critical angle, it will be the first to undergo total internal reflection as the angle of incidence is increased.

Question 16

A sound wave with frequency $f = 440\text{ Hz}$ travels from air into a medium where its speed increases to $1.5$ times the speed in air. A second sound wave with the same frequency travels from air into a different medium where its wavelength becomes $0.8$ times the wavelength in air. When these two transmitted waves meet at a boundary, what is the ratio of their wavelengths?

$1.2$ because the wavelength ratios are determined by the speed ratios in each medium
$1.875$ because wavelength scales with speed in the first medium but changes differently in the second (correct answer)
$0.8$ because both waves originated with the same frequency and wavelength relationships are preserved
$1.0$ because frequency is conserved across boundaries, making wavelengths equal in any common medium

Explanation: For the first wave: v₁ = 1.5v_air, and since f is constant, λ₁ = v₁/f = 1.5(v_air/f) = 1.5λ_air. For the second wave: λ₂ = 0.8λ_air is given directly. Therefore, λ₁/λ₂ = (1.5λ_air)/(0.8λ_air) = 1.5/0.8 = 1.875.

Question 17

Unpolarized light with intensity $I_0$ passes through three polarizing filters. The first filter has its transmission axis at $0°$ , the second at $45°$ , and the third at $90°$ . What is the intensity of light emerging from the third filter?

$\frac{I_0}{8}$ because each filter reduces intensity according to Malus's law with the cumulative angle differences (correct answer)
$0$ because the first and third filters have perpendicular transmission axes, blocking all light completely
$\frac{I_0}{4}$ because the intermediate filter allows some light to pass between the crossed polarizers
$\frac{I_0}{16}$ because unpolarized light requires special consideration with multiple sequential polarizations

Explanation: Unpolarized light through first polarizer: I₁ = I₀/2. Through second polarizer (45° relative to first): I₂ = I₁cos²(45°) = (I₀/2)(1/2) = I₀/4. Through third polarizer (45° relative to second): I₃ = I₂cos²(45°) = (I₀/4)(1/2) = I₀/8. The intermediate polarizer enables transmission between crossed polarizers.

Question 18

A water wave with amplitude $A = 0.20\text{ m}$ and wavelength $\lambda = 3.0\text{ m}$ travels at speed $v = 1.5\text{ m/s}$ . The wave encounters a barrier with a circular opening of diameter $d = 1.5\text{ m}$ . What type of wave behavior will be most prominent beyond the opening?

Refraction effects dominating because the circular geometry changes the effective wave speed through the opening
Minimal diffraction with mostly straight-line propagation because the opening is larger than the wavelength
Total reflection because the opening size creates impedance mismatch with the incident wave
Significant diffraction because the opening size is comparable to the wavelength, causing wave spreading (correct answer)

Explanation: When a wave encounters an opening or obstacle, the key factor determining the behavior is the relationship between the opening size and the wavelength. This is a classic diffraction scenario where you need to compare these two measurements. Here, the wavelength is

\lambda = 3.0\text{ m}

and the opening diameter is

d = 1.5\text{ m}

. Since the opening size is about half the wavelength (comparable in magnitude), significant diffraction will occur. When waves pass through openings similar in size to their wavelength, they spread out considerably beyond the opening rather than continuing in straight lines. Option D correctly identifies this diffraction effect. The opening size being comparable to the wavelength means the waves will bend and spread as they pass through, creating the characteristic semicircular wave pattern beyond the barrier. Option A incorrectly suggests refraction dominates. Refraction occurs when waves change speed between different media, but there's no indication of a medium change here - just a geometric opening in a barrier. Option B misapplies the diffraction principle. While the opening is indeed larger than the wavelength, it's not significantly larger. The rule of thumb is that when the opening is many times larger than the wavelength, you get minimal diffraction. Here, with

d ≈ 0.5\lambda

, diffraction effects are still prominent. Option C incorrectly invokes reflection and impedance mismatch. These concepts apply to boundaries between different media, not to geometric openings of this type. Study tip: Remember that significant diffraction occurs when the obstacle or opening size is comparable to (within a few times) the wavelength. If opening >> wavelength, minimal diffraction; if opening ≈ wavelength, significant diffraction.

Question 19

A guitar string vibrates in its third harmonic mode with frequency $f_3 = 660\text{ Hz}$ . The string length is $L = 0.65\text{ m}$ and linear mass density is $\mu = 3.2 \times 10^{-4}\text{ kg/m}$ . If the tension in the string is suddenly increased by $44\%$ , what will be the new frequency of the third harmonic?

$750\text{ Hz}$ because frequency increases proportionally with tension increase
$950\text{ Hz}$ because both tension and harmonic number effects must be considered together
$792\text{ Hz}$ because wave speed increases with the square root of tension (correct answer)
$660\text{ Hz}$ because the harmonic frequency ratios are independent of string tension

Explanation: When you encounter guitar string harmonics problems, you're dealing with standing waves on a fixed string, where both wave speed and harmonic frequencies are involved. The key insight is understanding how tension affects wave speed, which then affects frequency. For a guitar string, the wave speed is

v = \sqrt{\frac{T}{\mu}}

, where

T

is tension and

\mu

is linear mass density. The third harmonic frequency is

f_3 = \frac{3v}{2L}

. When tension increases by 44%, the new tension becomes

T_{new} = 1.44T

, so the new wave speed is

v_{new} = \sqrt{\frac{1.44T}{\mu}} = 1.2\sqrt{\frac{T}{\mu}} = 1.2v

. Therefore, the new third harmonic frequency is

f_{3,new} = \frac{3(1.2v)}{2L} = 1.2f_3 = 1.2 \times 660 = 792\text{ Hz}

. Option A incorrectly assumes frequency increases proportionally with tension, but frequency actually increases with the square root of tension due to the wave speed relationship. Option B mentions "harmonic number effects," but the harmonic number doesn't change in this problem—we're still looking at the third harmonic. Option D incorrectly claims harmonic frequencies are independent of tension, but since wave speed depends on tension, all harmonic frequencies will change when tension changes. Remember: for string instruments, frequency changes follow the square root relationship with tension. When tension increases by a factor, frequency increases by the square root of that factor—this appears frequently on IB Physics exams.

Question 20

A monochromatic light beam with wavelength $\lambda = 600\text{ nm}$ passes through two parallel slits separated by distance $d = 0.50\text{ mm}$ . The interference pattern is observed on a screen located $L = 2.0\text{ m}$ from the slits. If the entire apparatus is then submerged in water (refractive index $n = 1.33$ ), what happens to the fringe spacing?

The fringe spacing decreases by a factor of 1.33 because the wavelength in water decreases (correct answer)
The fringe spacing increases by a factor of 1.33 because the path difference requirements change in the medium
The fringe spacing remains unchanged because only the phase relationships matter, not the absolute wavelength
The fringe spacing decreases by a factor of 1.77 because both wavelength and effective slit separation change

Explanation: When light enters water, its wavelength decreases by a factor of n: λ_water = λ_air/n = 600nm/1.33 ≈ 451nm. The fringe spacing is given by Δy = λL/d, so when λ decreases by factor 1.33, the fringe spacing also decreases by factor 1.33. The slit separation d and screen distance L are physical distances that don't change.