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AP Physics 2 : Optics

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Example Questions

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Example Question #11 : Snell's Law

Light travels from air n=1 to crown glass n=1.52 . If the wavelength of incident light is 530nm , determine the wavelength of refracted light.

Possible Answers:

655nm

410nm

None of these

309nm

349nm

Correct answer:

349nm

Explanation:

Snell's Law:

$\frac{sin\theta_1}{sin\theta_2}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Where

$\theta$ is the respective incident angle

is the respective velocity of the light

$\lambda$ is the respective wavelength

is the respective index of refraction

Snells law for varsity

Use Snell's Law

$\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Solve for $\lambda_2$ :

$\lambda_2=\frac{n_1}{n_2}\lambda_1$

Plug in values:

$\lambda_2=\frac{1}{1.52}530nm$

$\lambda_2=349nm$

Report an Error

Example Question #12 : Snell's Law

Light travels from air n=1 to crown glass n=1.52 . If the wavelength of incident light is 690nm , determine the wavelength of refracted light.

Possible Answers:

None of these

822nm

475nm

454nm

432nm

Correct answer:

454nm

Explanation:

Snell's Law:

$\frac{sin\theta_1}{sin\theta_2}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Where

$\theta$ is the respective incident angle

is the respective velocity of the light

$\lambda$ is the respective wavelength

is the respective index of refraction

Snells law for varsity

Use Snell's Law:

$\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Solve for $\lambda_2$ :

$\lambda_2=\frac{n_1}{n_2}\lambda_1$

Plug in values:

$\lambda_2=\frac{1}{1.52}690nm$

$\lambda_2=454nm$

Report an Error

Example Question #13 : Snell's Law

Light in a vacuum hits the surface of an unknown material with an angle of incidence of $75^\circ$ . The angle of refraction is $23^\circ$ . What is the index of refraction of the unknown material?

Possible Answers:

3.38

3.88

None of these

2.47

1.41

Correct answer:

2.47

Explanation:

Snell's Law:

$\frac{sin\theta_1}{sin\theta_2}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Where

$\theta$ is the respective incident angle

is the respective velocity of the light

$\lambda$ is the respective wavelength

is the respective index of refraction

Snells law for varsity

Use Snell's Law:

$\frac{sin\theta_1}{sin\theta_2}=\frac{n_2}{n_1}$

Solve for n_2 :

$n_1\frac{sin\theta_1}{sin\theta_2}=n_2$

Plug in values:

$1*\frac{sin75^\circ}{sin23^\circ}=n_2$

2.47=n_2

Report an Error

Example Question #11 : Snell's Law

Light in a vacuum hits the surface of an unknown material with an angle of incidence of $35^\circ$ . The angle of refraction is $22.5^\circ$ . What is the index of refraction of the unknown material?

Possible Answers:

1.75

1.13

1.23

1.50

None of these

Correct answer:

1.50

Explanation:

Snell's Law:

$\frac{sin\theta_1}{sin\theta_2}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Where

$\theta$ is the respective incident angle

is the respective velocity of the light

$\lambda$ is the respective wavelength

is the respective index of refraction

Snells law for varsity

Use Snell's Law

$\frac{sin\theta_1}{sin\theta_2}=\frac{n_2}{n_1}$

Solve for n_2 :

$n_1\frac{sin\theta_1}{sin\theta_2}=n_2$

Plug in values:

$1*\frac{sin35^\circ}{sin22.5^\circ}=n_2$

1.50=n_2

Report an Error

Example Question #15 : Snell's Law

Light in a vacuum hits the surface of an unknown material with an angle of incidence of $47^\circ$ . The angle of refraction is $18^\circ$ . What is the index of refraction of the unknown material?

Possible Answers:

2.36

2.13

1.90

1.51

None of these

Correct answer:

2.36

Explanation:

Snell's Law:

$\frac{sin\theta_1}{sin\theta_2}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Where

$\theta$ is the respective incident angle

is the respective velocity of the light

$\lambda$ is the respective wavelength

is the respective index of refraction

Snells law for varsity

Use Snell's Law

$\frac{sin\theta_1}{sin\theta_2}=\frac{n_2}{n_1}$

Solve for n_2 :

$n_1\frac{sin\theta_1}{sin\theta_2}=n_2$

Plug in values:

$1*\frac{sin47^\circ}{sin18^\circ}=n_2$

2.36=n_2

Report an Error

Example Question #41 : Optics

Light travels from air n=1 to crown glass n=1.52 . If the wavelength of incident light is 620nm , determine the wavelength of refracted light.

Possible Answers:

591nm

435nm

408nm

467nm

None of these

Correct answer:

408nm

Explanation:

Snell's Law:

$\frac{sin\theta_1}{sin\theta_2}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Where

$\theta$ is the respective incident angle

is the respective velocity of the light

$\lambda$ is the respective wavelength

is the respective index of refraction

Snells law for varsity

Use Snell's Law

$\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Solve for $\lambda_2$ :

$\lambda_2=\frac{n_1}{n_2}\lambda_1$

Plug in values:

$\lambda_2=\frac{1}{1.52}620nm$

$\lambda_2=408nm$

Report an Error

Example Question #17 : Snell's Law

Light travels from air n=1 to crown glass n=1.52 . If the angle of incidence is $45^\circ$ , determine the angle of refraction.

Possible Answers:

$45^\circ$

$23.5^\circ$

$17.7^\circ$

$27.7^\circ$

$21.1^\circ$

Correct answer:

$27.7^\circ$

Explanation:

Snell's law:

$\frac{sin\theta_1}{sin\theta_2}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Where

$\theta$ is the respective incident angle

is the respective velocity of the light

$\lambda$ is the respective wavelength

is the respective index of refraction

Snells law for varsity fixed

Use Snell's law

$\frac{sin\theta_1}{sin\theta_2}=\frac{n_2}{n_1}$

Solve for $\theta_2$ :

$sin^{-1}(\frac{n_1 sin\theta_1^\circ}{n_2})=\theta_2$

Plug in values:

$sin^{-1}(\frac{1*sin45^\circ}{1.52})=\theta_2$

$27.7^\circ=\theta_2$

Report an Error

Example Question #11 : Snell's Law

Light travels from air n=1 to crown glass n=1.52 . If the angle of incidence is $22^\circ$ , determine the angle of refraction.

Possible Answers:

$14.3^\circ$

$17.3^\circ$

$10.8^\circ$

$21.5^\circ$

$9.3^\circ$

Correct answer:

$14.3^\circ$

Explanation:

Snell's law:

$\frac{sin\theta_1}{sin\theta_2}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Where

$\theta$ is the respective incident angle

is the respective velocity of the light

$\lambda$ is the respective wavelength

is the respective index of refraction

Snells law for varsity fixed

Use Snell's law

$\frac{sin\theta_1}{sin\theta_2}=\frac{n_2}{n_1}$

Solve for $\theta_2$ :

$sin^{-1}(\frac{n_1 sin\theta_1^\circ}{n_2})=\theta_2$

Plug in values:

$sin^{-1}(\frac{1*sin22^\circ}{1.52})=\theta_2$

$14.3^\circ=\theta_2$

Report an Error

Example Question #19 : Snell's Law

Light travels from air n=1 to crown glass n=1.52 . If the angle of incidence is $78^\circ$ , determine the angle of refraction.

Possible Answers:

$11.6^\circ$

$37.7^\circ$

$40.1^\circ$

$34.2^\circ$

$45.2^\circ$

Correct answer:

$40.1^\circ$

Explanation:

Snell's law:

$\frac{sin\theta_1}{sin\theta_2}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Where

$\theta$ is the respective incident angle

is the respective velocity of the light

$\lambda$ is the respective wavelength

is the respective index of refraction

Snells law for varsity fixed

Use

$\frac{sin\theta_1}{sin\theta_2}=\frac{n_2}{n_1}$

Solve for $\theta_2$ :

$sin^{-1}(\frac{n_1 sin\theta_1^\circ}{n_2})=\theta_2$

Plug in values:

$sin^{-1}(\frac{1*sin78^\circ}{1.52})=\theta_2$

$40.1^\circ=\theta_2$

Report an Error

Example Question #11 : Snell's Law

Light travels from air n=1 to water n=1.33 . If the wavelength of incident light is 530nm , determine the wavelength of refracted light.

Possible Answers:

457nm

320nm

251nm

398nm

None of these

Correct answer:

398nm

Explanation:

Snell's law:

$\frac{sin\theta_1}{sin\theta_2}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Where

$\theta$ is the respective incident angle

is the respective velocity of the light

$\lambda$ is the respective wavelength

is the respective index of refraction

Snells law for varsity fixed

Use Snell's law:

$\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}$

Solve for $\lambda_2$

$\lambda_2=\frac{n_1}{n_2}\lambda_1$

Plug in values:

$\lambda_2=\frac{1}{1.33}530nm$

$\lambda_2=398nm$

Report an Error

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AP Physics 2 : Optics

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #11 : Snell's Law

Example Question #12 : Snell's Law

Example Question #13 : Snell's Law

Example Question #11 : Snell's Law

Example Question #15 : Snell's Law

Example Question #41 : Optics

Example Question #17 : Snell's Law

Example Question #11 : Snell's Law

Example Question #19 : Snell's Law

Example Question #11 : Snell's Law

All AP Physics 2 Resources

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