What is Waves & light on the Physics Regents?

Waves & light is a Waves topic on the Physics Regents. RegentsLab has 52 practice questions on it, each with a full worked solution.

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Waves & light: Physics Regents Practice

Practice Waves & light for the Physics Regents with real-style questions and full worked solutions. Miss one? Read the solution — the exam reuses the same ideas every year.

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Question 65 · 2 pts

A periodic wave has a frequency of $50$ Hz and a wavelength of $4.0$ m. What is the speed of the wave?

(A) $54$ m/s
(B) $0.08$ m/s
(C) $12.5$ m/s
(D) $200$ m/s ✓

Worked solution

$v = f\lambda$ Wave-speed equation from the reference tables

$v = (50\ \text{Hz})(4.0\ \text{m}) = 200\ \text{m/s}$

Answer: $200$ m/s

The speed of any periodic wave equals its frequency multiplied by its wavelength, so larger frequency or longer wavelength means a faster wave.

Question 66 · 2 pts

A sound wave has a frequency of $250$ Hz. What is its period?

(A) $4.0\times10^{-3}$ s ✓
(B) $4.0\times10^{-2}$ s
(C) $2.0\times10^{-3}$ s
(D) $2.5\times10^{2}$ s

Worked solution

$T = \dfrac{1}{f}$ Period is the reciprocal of frequency

$T = \dfrac{1}{250\ \text{Hz}} = 4.0\times10^{-3}\ \text{s}$

Answer: $4.0\times10^{-3}$ s

Period and frequency are reciprocals of each other, so a high frequency corresponds to a short period.

Question 75 · 2 pts

A train moving at constant speed sounds a whistle of constant frequency as it approaches a stationary observer. Compared to the sound emitted by the whistle, the sound detected by the observer has a

(A) lower frequency and longer wavelength
(B) higher frequency and longer wavelength
(C) higher frequency and shorter wavelength ✓
(D) same frequency and same wavelength

Worked solution

$\text{Approaching source} \Rightarrow f_{\text{detected}} \uparrow$ The wavefronts bunch up in front of the moving source.

$v = f\lambda \Rightarrow \lambda = \dfrac{v}{f}$ The speed of sound $v$ is constant, so a higher $f$ forces a shorter $\lambda$.

$f \uparrow \Rightarrow \lambda \downarrow$

Answer: higher frequency and shorter wavelength

As a wave source approaches an observer the waves pile up, so the detected frequency rises and the detected wavelength shortens, which is the Doppler effect.

Question 79 · 2 pts

A periodic transverse wave has a wavelength of $0.40$ m. Two points on this wave vibrate exactly in phase with each other when they are separated by a distance of

(A) $0.10$ m
(B) $0.20$ m
(C) $0.60$ m
(D) $0.80$ m ✓

Worked solution

$\text{In phase} \Rightarrow \text{separation} = n\lambda,\quad n = 1, 2, 3, \dots$ Only whole-number multiples of a wavelength are in phase.

$2\lambda = 2(0.40\ \text{m}) = 0.80\ \text{m}$

Answer: $0.80$ m

Points on a wave are in phase only when separated by a whole number of wavelengths; separations of $\lambda/2$ or $1.5\lambda$ leave points exactly out of phase.

Question 9 · 2 pts

A wave has a wavelength of $2.0\ \text{m}$ and a frequency of $5.0\ \text{Hz}$. What is the speed of the wave?

(A) $0.40\ \text{m/s}$
(B) $7.0\ \text{m/s}$
(C) $2.5\ \text{m/s}$
(D) $10\ \text{m/s}$ ✓

Worked solution

$v = f\lambda$ Universal wave equation

$v = (5.0\ \text{Hz})(2.0\ \text{m})$

$v = 10\ \text{m/s}$

Answer: $10\ \text{m/s}$

Wave speed equals frequency times wavelength; multiplying the given values gives 10 m/s.

Question 22 · 2 pts

A wave traveling in a uniform medium has a frequency of $5.0\text{ Hz}$ and a wavelength of $2.0\text{ m}$. What is the speed of the wave?

(A) $7.0\text{ m/s}$
(B) $2.5\text{ m/s}$
(C) $0.40\text{ m/s}$
(D) $10.\text{ m/s}$ ✓

Worked solution

$v = f\lambda$ Universal wave equation

$v = (5.0\text{ Hz})(2.0\text{ m})$

$v = 10.\text{ m/s}$

Answer: $10.\text{ m/s}$

The speed of a periodic wave equals the product of its frequency and wavelength, giving 10. m/s.

Question 48 · 2 pts

A wave traveling in a uniform medium has a frequency of 500. Hz and a wavelength of 0.60 m. What is the speed of the wave?

(A) 0.0012 m/s
(B) 300.6 m/s
(C) 300 m/s ✓
(D) 833 m/s

Worked solution

$v = f \lambda$

$v = (500.\ \text{Hz})(0.60\ \text{m})$

$v = 300\ \text{m/s}$

Answer: 300 m/s

Wave speed equals frequency multiplied by wavelength; multiplying 500. Hz by 0.60 m gives 300 m/s. Dividing instead of multiplying, or adding the two values, produces the incorrect distractors.

Question 23 · 2 pts

A wave traveling in a uniform medium has a frequency of 5.0 Hz and a wavelength of 2.0 m. What is the speed of the wave?

(A) 7.0 m/s
(B) 20 m/s
(C) 10 m/s ✓
(D) 2.5 m/s

Worked solution

$v = f\lambda$

$v = (5.0\ \text{Hz})(2.0\ \text{m})$

$v = 10\ \text{m/s}$

Answer: 10 m/s

The universal wave equation relates speed, frequency, and wavelength by $v=f\lambda$. Dividing instead of multiplying the two quantities (giving 2.5 m/s) is a common student error.

Question 46 · 2 pts

A sound wave has a frequency of 500 Hz and a wavelength of 0.68 m. What is the speed of this wave?

(A) 500 m/s
(B) 0.0014 m/s
(C) 735 m/s
(D) 340 m/s ✓

Worked solution

$v = f\lambda$

$v = (500\text{ Hz})(0.68\text{ m})$

$v = 340\text{ m/s}$

Answer: 340 m/s

Wave speed equals the product of frequency and wavelength; multiplying 500 Hz by 0.68 m gives 340 m/s, which is the well-known approximate speed of sound in air.

Question 57 · 4 pts · constructed response

A sound wave has a wavelength of $0.50\ \text{m}$ and a frequency of $660\ \text{Hz}$. Calculate the speed of the wave and its period.

Model answer: The wave speed is 330 m/s, and its period is about 1.5 x 10^-3 s.

Worked solution

$v = f\lambda$ Universal wave equation

$v = (660)(0.50) = 330\ \text{m/s}$

$T = \dfrac{1}{f} = \dfrac{1}{660} = 1.5\times10^{-3}\ \text{s}$

Answer: The wave speed is 330 m/s, and its period is about 1.5 x 10^-3 s.

The wave equation relates speed, frequency, and wavelength, while the period is simply the reciprocal of the frequency.

Question 10 · 2 pts

A wave has a frequency of $200\ \text{Hz}$ and a wavelength of $1.5\ \text{m}$. What is the speed of the wave?

(A) $0.0075\ \text{m/s}$
(B) $300\ \text{m/s}$ ✓
(C) $201.5\ \text{m/s}$
(D) $133\ \text{m/s}$

Worked solution

$v = f\lambda$ universal wave equation

$v = (200\ \text{Hz})(1.5\ \text{m})$

$v = 300\ \text{m/s}$

Answer: $300\ \text{m/s}$

The universal wave equation states that wave speed equals frequency times wavelength; multiplying the given values gives 300 m/s.

Question 24 · 2 pts

A wave has a frequency of $200.\ \text{Hz}$ and a wavelength of $1.5\ \text{m}$. What is the speed of the wave?

(A) $2.0 \times 10^2\ \text{m/s}$
(B) $2.0 \times 10^1\ \text{m/s}$
(C) $3.0 \times 10^2\ \text{m/s}$ ✓
(D) $1.3 \times 10^2\ \text{m/s}$

Worked solution

$v = f\lambda$ universal wave equation

$v = (200.\ \text{Hz})(1.5\ \text{m})$

$v = 3.0 \times 10^2\ \text{m/s}$

Answer: $3.0 \times 10^2\ \text{m/s}$

The speed of any wave equals the product of its frequency and wavelength; multiplying the given values gives 300 m/s.

Question 25 · 2 pts

A sound wave traveling through air has a frequency of $500.\ \text{Hz}$ and travels at a speed of $340\ \text{m/s}$. What is the wavelength of this sound wave?

(A) 0.68 m ✓
(B) 170 m
(C) 170,000 m
(D) 1.5 m

Worked solution

$v = f\lambda \Rightarrow \lambda = \dfrac{v}{f}$ wave speed equation

$\lambda = \dfrac{340\ \text{m/s}}{500.\ \text{Hz}} = 0.68\ \text{m}$

Answer: 0.68 m

The universal wave equation relates speed, frequency, and wavelength as $v = f\lambda$. Solving for wavelength gives the distance between successive wave crests for this sound wave.

Question 48 · 2 pts

A wave has a frequency of 5.0 hertz and a wavelength of 2.0 meters. What is the speed of the wave?

(A) 10 m/s ✓
(B) 2.5 m/s
(C) 7.0 m/s
(D) 0.40 m/s

Worked solution

$v = f\lambda = (5.0)(2.0) = 10\ \text{m/s}$ Wave speed equation

Answer: 10 m/s

Wave speed equals frequency times wavelength, giving 10 m/s.

Question 55 · 4 pts · constructed response

A sound wave traveling through air has a frequency of $440\ \text{Hz}$ and a wavelength of $0.78\ \text{m}$. Calculate the speed of the sound wave.

Model answer: The speed of the sound wave is 343 m/s.

Worked solution

$v = f\lambda$ universal wave equation

$v = (440\ \text{Hz})(0.78\ \text{m})$

$v = 343.2\ \text{m/s} \approx 343\ \text{m/s}$

Answer: The speed of the sound wave is 343 m/s.

The speed of any periodic wave equals the product of its frequency and wavelength, giving a value consistent with the known speed of sound in air, about 343 m/s.

Question 24 · 1 pt

A wave has a frequency of $5\ \mathrm{Hz}$ and a wavelength of $2\ \mathrm{m}$. What is its speed?

(A) $7\ \mathrm{m/s}$
(B) $10\ \mathrm{m/s}$ ✓
(C) $0.4\ \mathrm{m/s}$
(D) $2.5\ \mathrm{m/s}$

Worked solution

$v = f\lambda = (5)(2)$ Wave-speed equation.

$= 10\ \mathrm{m/s}$

Answer: 10\ \mathrm{m/s}

Wave speed equals frequency times wavelength (v = fλ).

Question 67 · 2 pts

Yellow light travels through a vacuum with a wavelength of $5.9\times10^{-7}$ m. What is the frequency of this light? (Use $c = 3.00\times10^{8}$ m/s.)

(A) $2.0\times10^{-15}$ Hz
(B) $5.1\times10^{14}$ Hz ✓
(C) $5.1\times10^{15}$ Hz
(D) $1.8\times10^{2}$ Hz

Worked solution

$c = f\lambda \Rightarrow f = \dfrac{c}{\lambda}$ Light obeys v = f\lambda with v = c in a vacuum

$f = \dfrac{3.00\times10^{8}}{5.9\times10^{-7}} = 5.1\times10^{14}\ \text{Hz}$

Answer: $5.1\times10^{14}$ Hz

All electromagnetic waves travel at $c$ in a vacuum, so frequency is found by dividing the speed of light by the wavelength.

Question 68 · 2 pts

Light travels through a transparent material at a speed of $2.0\times10^{8}$ m/s. What is the absolute index of refraction of the material? (Use $c = 3.00\times10^{8}$ m/s.)

(A) $0.67$
(B) $1.5\times10^{8}$
(C) $6.0\times10^{16}$
(D) $1.5$ ✓

Worked solution

$n = \dfrac{c}{v}$ Absolute index of refraction from the reference tables

$n = \dfrac{3.00\times10^{8}}{2.0\times10^{8}} = 1.5$

Answer: $1.5$

The index of refraction is the ratio of the speed of light in a vacuum to its speed in the medium, so it is always greater than or equal to 1 and has no units.

Question 69 · 2 pts

A ray of light traveling in air ($n = 1.00$) strikes the surface of water ($n = 1.33$) at an angle of incidence of $30.^\circ$. What is the angle of refraction in the water?

(A) $42^\circ$
(B) $0.38^\circ$
(C) $22^\circ$ ✓
(D) $30^\circ$

Worked solution

$n_1 \sin\theta_1 = n_2 \sin\theta_2$ Snell's law

$\sin\theta_2 = \dfrac{(1.00)\sin 30^\circ}{1.33} = 0.376$

$\theta_2 = \sin^{-1}(0.376) = 22^\circ$

Answer: $22^\circ$

When light enters a medium with a higher index of refraction it slows down and bends toward the normal, giving a smaller angle of refraction.

Question 73 · 2 pts

Monochromatic light with a wavelength of $5.0 \times 10^{-7}$ m travels through air ($n = 1.00$) and then enters water ($n = 1.33$). As the light enters the water its frequency stays the same. What is the wavelength of this light in the water?

(A) $6.7 \times 10^{-7}$ m
(B) $5.0 \times 10^{-7}$ m
(C) $2.8 \times 10^{-7}$ m
(D) $3.8 \times 10^{-7}$ m ✓

Worked solution

$\lambda_{\text{medium}} = \dfrac{\lambda_{\text{air}}}{n}$ Frequency is constant, so with $v = c/n$ and $v = f\lambda$ the wavelength scales as $1/n$.

$\lambda_{\text{water}} = \dfrac{5.0 \times 10^{-7}\ \text{m}}{1.33}$

$\lambda_{\text{water}} = 3.8 \times 10^{-7}\ \text{m}$

Answer: $3.8 \times 10^{-7}$ m

When light enters a denser medium its speed and wavelength decrease in proportion to the index of refraction, while its frequency and color stay the same.

Question 77 · 2 pts

A vibrating source produces a periodic wave with a frequency of $5.0$ Hz. The wave travels a distance of $12$ m in $4.0$ s. What is the wavelength of the wave?

(A) $0.60$ m ✓
(B) $2.4$ m
(C) $15$ m
(D) $1.7$ m

Worked solution

$v = \dfrac{d}{t} = \dfrac{12\ \text{m}}{4.0\ \text{s}} = 3.0\ \text{m/s}$

$v = f\lambda \Rightarrow \lambda = \dfrac{v}{f}$

$\lambda = \dfrac{3.0\ \text{m/s}}{5.0\ \text{Hz}} = 0.60\ \text{m}$

Answer: $0.60$ m

Wave speed comes from $v = d/t$, then the wavelength follows from $v = f\lambda$; the speed is set by the medium, not by how fast the source vibrates.

Question 78 · 2 pts

A ray of light traveling in air strikes the flat surface of a pool of water ($n = 1.33$) with an angle of incidence of $30.^\circ$ measured from the normal. What is the angle of refraction in the water?

(A) $22^\circ$ ✓
(B) $42^\circ$
(C) $30^\circ$
(D) $68^\circ$

Worked solution

$n_1\sin\theta_1 = n_2\sin\theta_2$

$\sin\theta_2 = \dfrac{(1.00)\sin 30^\circ}{1.33} = \dfrac{0.500}{1.33} = 0.376$

$\theta_2 = \sin^{-1}(0.376) = 22^\circ$ Light entering a denser medium bends toward the normal, so $\theta_2 < \theta_1$.

Answer: $22^\circ$

Snell's law ($n_1\sin\theta_1 = n_2\sin\theta_2$) shows light slows and bends toward the normal entering a higher-index medium, with all angles measured from the normal.

Question 80 · 2 pts

Two pulses travel toward each other along the same rope. One pulse has a maximum displacement of $0.30$ m above the rope's rest position, and the other has a maximum displacement of $0.20$ m below it. At the instant the two pulses completely overlap, what is the displacement of the rope at that point?

(A) $0.0$ m
(B) $0.50$ m
(C) $0.25$ m
(D) $0.10$ m ✓

Worked solution

$y_{\text{net}} = y_1 + y_2$ By superposition, displacements add algebraically with above as $+$ and below as $-$.

$y_{\text{net}} = (+0.30\ \text{m}) + (-0.20\ \text{m})$

$y_{\text{net}} = +0.10\ \text{m}$

Answer: $0.10$ m

The principle of superposition makes the displacements of overlapping pulses add algebraically, so opposite-side pulses partially cancel while same-side pulses reinforce.

Question 10 · 2 pts

Light traveling in a vacuum at $3.00\times10^{8}\ \text{m/s}$ enters a block of glass with an index of refraction of $1.5$. What is the speed of light inside the glass?

(A) $4.50\times10^{8}\ \text{m/s}$
(B) $1.50\times10^{8}\ \text{m/s}$
(C) $3.00\times10^{8}\ \text{m/s}$
(D) $2.00\times10^{8}\ \text{m/s}$ ✓

Worked solution

$n = \dfrac{c}{v} \Rightarrow v = \dfrac{c}{n}$ Definition of index of refraction

$v = \dfrac{3.00\times10^{8}\ \text{m/s}}{1.5}$

$v = 2.00\times10^{8}\ \text{m/s}$

Answer: $2.00\times10^{8}\ \text{m/s}$

The index of refraction is the ratio of the speed of light in a vacuum to its speed in the medium; dividing c by n gives the reduced speed of light in the glass.

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