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AP Physics 2 : AP Physics 2

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

Example Questions

Example Question #471 : Ap Physics 2

Light travels from a vacuum, n=1 , to a material with n=1.74 . What will be the velocity in the new medium?

Possible Answers:

$1.72*10^8\frac{m}{s}$

None of these

$1.92*10^8\frac{m}{s}$

$1.22*10^8\frac{m}{s}$

$1.58*10^8\frac{m}{s}$

Correct answer:

$1.72*10^8\frac{m}{s}$

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:

$\frac{v_1}{v_2}=\frac{n_2}{n_1}$

Solve for v_2

$\frac{n_1}{n_2}v_1=v_2$

The velocity of light in a vacuum:

$v_1=c=3.0*10^8\frac{m}{s}$

Plug in values:

$\frac{1}{1.74}3.0*10^8\frac{m}{s}=v_2$

$1.72*10^8\frac{m}{s}=v_2$

Report an Error

Example Question #21 : Snell's Law

Light travels from a vacuum, n=1 , to a material with n=1.56 . What will be the velocity in the new medium?

Possible Answers:

$3.00*10^8\frac{m}{s}$

$4.5*10^8\frac{m}{s}$

$1.92*10^8\frac{m}{s}$

None of these

$1.55*10^8\frac{m}{s}$

Correct answer:

$1.92*10^8\frac{m}{s}$

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:

$\frac{v_1}{v_2}=\frac{n_2}{n_1}$

Solve for v_2

$\frac{n_1}{n_2}v_1=v_2$

The velocity of light in a vacuum,

$v_1=c=3.0*10^8\frac{m}{s}$

Plug in values:

$\frac{1}{1.56}3.0*10^8\frac{m}{s}=v_2$

$1.92*10^8\frac{m}{s}=v_2$

Report an Error

Example Question #22 : Snell's Law

Light travels from water n=1.33 to diamond n=2.42 . If the angle of incidence was $32^\circ$ , determine the angle of refraction.

Possible Answers:

$16.9^\circ$

$14.7^\circ$

$6.2^\circ$

$26.2^\circ$

None of these

Correct answer:

$16.9^\circ$

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:

$sin^{-1}(\frac{1.33*sin32^\circ}{2.42})=\theta_2$

$16.9^\circ=\theta_2$

Report an Error

Example Question #21 : Snell's Law

Light of wavelength 450nm strikes an interface of transparent materials at and angle of incidence of $35^{\circ}$ . The angle of refraction is $22^{\circ}$ . Determine the wavelength of the light in the new material.

Possible Answers:

701nm

294nm

565nm

212nm

354nm

Correct answer:

294nm

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:

$\frac{sin\theta_1}{sin\theta_2}=\frac{\lambda_1}{\lambda_2}$

Plug in values:

$\frac{sin22^\circ}{sin35^\circ}=\frac{\lambda_1}{450nm}$

Solve for $\lambda_1$ :

$294nm=\lambda_1$

Report an Error

Example Question #27 : Snell's Law

Light travels from water n=1.33 to diamond n=2.42 . If the wavelength of incident light was 532nm , determine the wavelength of refracted light.

Possible Answers:

292nm

390nm

420nm

766nm

357nm

Correct answer:

292nm

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:

$\lambda_2=\frac{1.33}{2.42}532nm$

$\lambda_2=292nm$

Report an Error

Example Question #21 : Snell's Law

Light travels from a vacuum n=1 to a material with n=1.84 . What will be the velocity in the new medium?

Possible Answers:

$1.44*10^8\frac{m}{s}$

$2.22*10^8\frac{m}{s}$

$1.63*10^8\frac{m}{s}$

$1.84*10^8\frac{m}{s}$

$1.11*10^8\frac{m}{s}$

Correct answer:

$1.63*10^8\frac{m}{s}$

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:

$\frac{v_1}{v_2}=\frac{n_2}{n_1}$

Solve for v_2 :

$\frac{n_1}{n_2}v_1=v_2$

The velocity of light in a vacuum is

$v_1=c=3.0*10^8\frac{m}{s}$

Plug in values:

$\frac{1}{1.84}3.0*10^8\frac{m}{s}=v_2$

$1.63*10^8\frac{m}{s}=v_2$

Report an Error

Example Question #25 : Snell's Law

Light travels from water (n=1.33) to an unknown transparent medium. The angle of incidence is $45^\circ$ . The angle of refraction is $55^{\circ}$ . Determine the speed of light in the new medium.

Possible Answers:

$2.9*10^8\frac{m}{s}$

$1.7*10^8\frac{m}{s}$

$2.6*10^8\frac{m}{s}$

$2.3*10^8\frac{m}{s}$

$8.1*10^8\frac{m}{s}$

Correct answer:

$2.6*10^8\frac{m}{s}$

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:

$\frac{v_3}{v_2}=\frac{n_2}{n_3}$

and

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

Where subscript represents water, subscript represents the unknown medium, and subscript represents vacuum

Plug in values:

$\frac{sin45^\circ}{sin55^\circ}=\frac{n_2}{1.33}$

Solve for n_2 :

n_2=1.15

$\frac{3.0*10^8\frac{m}{s}}{v_2}=\frac{1.15}{1}$

Solve for v_2 :

$v_2=2.6*10^8\frac{m}{s}$

Report an Error

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:

$17.2^\circ$

$7.1^\circ$

Impossible to determine

$8.3^\circ$

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 #62 : 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:

$333\textup{ nm}$

$110\textup{ nm}$

$285\textup{ nm}$

None of these

$450\textup{ nm}$

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 #63 : 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:

$18.9^\circ$

None of these

$23.5^\circ$

$56.0^\circ$

$47.5^\circ$

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

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All AP Physics 2 Resources

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AP Physics 2 : AP Physics 2

Study concepts, example questions & explanations for AP Physics 2

All AP Physics 2 Resources

Example Questions

Example Question #471 : Ap Physics 2

Example Question #21 : Snell's Law

Example Question #22 : Snell's Law

Example Question #21 : Snell's Law

Example Question #27 : Snell's Law

Example Question #21 : Snell's Law

Example Question #25 : Snell's Law

Example Question #61 : Optics

Example Question #62 : Optics

Example Question #63 : Optics

All AP Physics 2 Resources

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