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AP Physics 2 : Optics

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6 Diagnostic Tests 149 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #61 : Optics

Light travel from water n=1.33 to diamond n=2.42 . If the angle of incidence was $13^\circ$ , determine the angle of refraction.

Possible Answers:

$17.2^\circ$

$8.3^\circ$

Impossible to determine

$7.1^\circ$

None of these

Correct answer:

$7.1^\circ$

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 Snell's law:

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

Solving for $\theta_2$ :

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

Plugging in values:

$sin^{-1}(\frac{1.33*sin13^\circ}{2.42})=\theta_2$

$7.1^\circ=\theta_2$

Report an Error

Example Question #61 : Optics

Light travels from air n=1 to diamond n=2.42 . If the wavelength of incident light is $690 \textup{ nm}$ , determine the wavelength of refracted light.

Possible Answers:

$285\textup{ nm}$

$110\textup{ nm}$

$333\textup{ nm}$

$450\textup{ nm}$

None of these

Correct answer:

$285\textup{ nm}$

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

Using this version of Snell's law and solving for $\lambda_2$ :

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

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

Plugging in values:

$\lambda_2=\frac{1}{2.42}690nm$

$\lambda_2=285nm$

Report an Error

Example Question #62 : Optics

Light travels from air n=1 to diamond n=2.42 . If the angle of incidence is $75^\circ$ , determine the angle of refraction.

Possible Answers:

$18.9^\circ$

$47.5^\circ$

$56.0^\circ$

$23.5^\circ$

None of these

Correct answer:

$23.5^\circ$

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

Using the following version of Snell's law and solving for the angle of refraction:

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

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

Plugging in values:

$sin^{-1}(\frac{1*sin75^\circ}{2.42})=\theta_2$

$23.5^\circ=\theta_2$

Report an Error

Example Question #63 : Optics

Light in a vacuum hits the surface of an unknown material with an angle of incidence of $29^\circ$ . The angle of refraction is $10^\circ$ . What is the index of refraction of the unknown material?

Possible Answers:

2.45

2.79

2.32

None of these

1.89

Correct answer:

2.79

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

Using the following version of Snell's law and solving for n_2 :

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

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

Plugging in values

$1*\frac{sin29^\circ}{sin10^\circ}=n_2$

2.79=n_2

Report an Error

Example Question #62 : Optics

Light in a vacuum travels into an unknown medium. If the angle of incidence was $67^\circ$ and the angle of refraction is $37^\circ$ , determine the index of refraction of the new medium.

Possible Answers:

1.33

None of these

1.53

2.41

1.88

Correct answer:

1.53

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 the following version of Snell's law:

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

Solving for n_2

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

Plugging in values, remembering that for a vacuum, n=1

$1\frac{sin67^\circ}{sin37^\circ}=n_2$

n_2=1.53

Report an Error

Example Question #35 : Snell's Law

Light travels from air n=1 to diamond n=2.42 . If the wavelength of incident light is $720\textup{ nm}$ , determine the wavelength of refracted light.

Possible Answers:

$235\textup{ nm}$

None of these

$115\textup{ nm}$

$350\textup{ nm}$

$298\textup{ nm}$

Correct answer:

$298\textup{ nm}$

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 the following version of Snell's law:

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

Solving for $\lambda_2$

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

Plugging in values:

$\lambda_2=\frac{1}{2.42}720m$

$\lambda_2=298nm$

Report an Error

Example Question #37 : Snell's Law

Light travels from air n=1 to diamond n=2.42 . If the angle of incidence is $67^\circ$ , determine the angle of refraction.

Possible Answers:

$27.7^\circ$

$29.5^\circ$

$22.3^\circ$

$14.3^\circ$

None of these

Correct answer:

$22.3^\circ$

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 the following version of Snell's law:

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

Solving for $\theta_2$ :

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

Plugging in values:

$sin^{-1}(\frac{1*sin67^\circ}{2.42})=\theta_2$

$22.3^\circ=\theta_2$

Report an Error

Example Question #36 : Snell's Law

Light in a vacuum hits the surface of an unknown material with an angle of incidence of $88^\circ$ . The angle of refraction is $22.5^\circ$ . What is the index of refraction of the unknown material?

Possible Answers:

1.98

2.11

1.01

3.45

2.61

Correct answer:

2.61

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 the following version of Snell's law:

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

Solving for n_2

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

Plugging in values

$1*\frac{sin88^\circ}{sin22.5^\circ}=n_2$

2.61=n_2

Report an Error

Example Question #63 : Optics

Light travels from air n=1 to diamond n=2.42 . If the wavelength of incident light is $740\textup{ nm}$ , determine the wavelength of refracted light.

Possible Answers:

$306\textup{ nm}$

$490\textup{ nm}$

$452\textup{ nm}$

$595\textup{ nm}$

$390\textup{ nm}$

Correct answer:

$306\textup{ nm}$

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 the following version of Snell's law:

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

Solving for $\lambda_2$

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

Plugging in values:

$\lambda_2=\frac{1}{2.42}740m$

$\lambda_2=306nm$

Report an Error

Example Question #40 : Snell's Law

Light travels from air n=1 to diamond n=2.42 . If the angle of incidence is $87^\circ$ , determine the angle of refraction.

Possible Answers:

$47.4^\circ$

$35.4^\circ$

$24.4^\circ$

$26.8^\circ$

$29.9^\circ$

Correct answer:

$24.4^\circ$

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 the following version of Snell's law:

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

Solving for $\theta_2$ :

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

Plugging in values:

$sin^{-1}(\frac{1*sin87^\circ}{2.42})=\theta_2$

$24.4^\circ=\theta_2$

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 #61 : Optics

Example Question #61 : Optics

Example Question #62 : Optics

Example Question #63 : Optics

Example Question #62 : Optics

Example Question #35 : Snell's Law

Example Question #37 : Snell's Law

Example Question #36 : Snell's Law

Example Question #63 : Optics

Example Question #40 : Snell's Law

All AP Physics 2 Resources

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