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AP Physics 2 : Optics

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Example Question #41 : Snell's Law

Light in a vacuum hits the surface of an unknown material with an angle of incidence of $77^\circ$ . The angle of refraction is $68^\circ$ . What is the index of refraction of the unknown material?

Possible Answers:

1.46

1.32

1.25

1.17

1.05

Correct answer:

1.05

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

$1*\frac{sin77^\circ}{sin68^\circ}=n_2$

1.05=n_2

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Example Question #42 : Snell's Law

Suppose that a ray of light passes from air into another medium. The light ray strikes the surface of the medium at an angle of $30^{o}$ with respect to the normal. If the light ray is refracted such that it bends $15^{o}$ with respect to the normal, what is the index of refraction of this medium?

Note: The index of refraction of air is .

Possible Answers:

1.50

1.93

1.33

1.69

1.12

Correct answer:

1.93

Explanation:

In this question, we're told that a ray of light is hitting a medium which causes it to undergo refraction. We're provided with each of the angles with respect to the normal, as well as the index of refraction of the first medium. We're asked to find the index of refraction for the second medium.

First and foremost, we'll need to use the equation for Snell's law.

$n_{1}sin(\Theta _{1})=n_{2}sin(\Theta _{2})$

If we treat $n_{1}$ as the index of refraction for air, we can then use $n_{2}$ as the index of refraction for our unknown medium. Then, we can rearrange the equation to isolate the $n_{2}$ variable.

$n_{2}=n_{1}\frac{sin(\Theta _{1})}{sin(\Theta _{2})}$

Now, we can plug in the values that we know to solve for our answer.

$n_{2}=1\frac{sin(30^{o})}{sin(15^{o})}=1.93$

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Example Question #43 : Snell's Law

A beam of light in a vacuum hits a crystal with an angle of incidence of $45^\circ$ . The angle of refraction is determined to be $18^\circ$ , determine the index of refraction of the crystal

Possible Answers:

3.20

2.29

2.96

1.52

1.11

Correct answer:

2.29

Explanation:

Using Snell's Law:

$n_1sin\theta_1=n_2sin\theta_2$

The index of refraction of a vacuum is unity, n=1

Plugging in values:

$1sin45^\circ=n_2sin18^\circ$

n_2=2.29

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Example Question #71 : Optics

A photon moving in air enters a region with index of refraction $n_{r}$ . The region has length . At the end of the region, there's a mirror that reflects the light back through the region to the air. A detector measures that the photon is in the region for time . What is $n_{r}$ ? Let the speed of light be .

Possible Answers:

$\frac{2d}{t}$

$\frac{ct}{d}$

$\frac{ct}{2d}$

$\frac{2dc}{t}$

$\frac{t}{2d}$

Correct answer:

$\frac{ct}{2d}$

Explanation:

The index of refraction of a material is equal to $\frac{c}{v}$ , where is the speed of light in a vacuum and is the speed of light in the material. In the material given, the light travels a distance of (since it travels to the mirror and back) in a time . Speed is distance over time, so the light's speed in the material is $\frac{2d}{t}$ . Thus the index of refraction is

$\frac{c}{\frac{2d}{t}}$

Which when simplified, gives:

$\frac{ct}{2d}$

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Example Question #45 : Snell's Law

Light of wavelength 650nm goes from vacuum to an unknown liquid with an angle of incidence of $33^\circ$ and has an angle of diffraction of $15^\circ$ . Determine the index of refraction in the unknown liquid.

Possible Answers:

1.33

1.45

2.10

None of these

3.65

Correct answer:

2.10

Explanation:

Using Snell's law:

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

$\frac{1}{n_2}=\frac{sin15^\circ}{sin33^\circ}$

n_2=2.10

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Example Question #46 : Snell's Law

Light of wavelength 650nm goes from vacuum to an unknown liquid with an angle of incidence of $33^\circ$ and has an angle of diffraction of $15^\circ$ . Determine the wavelength in the unknown liquid.

Possible Answers:

488nm

1001nm

None of these

309nm

388nm

Correct answer:

309nm

Explanation:

Using Snell's law:

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

$\frac{sin 15^\circ}{sin 33^\circ}=\frac{\lambda_2}{650nm}$

$\lambda_2=309nm$

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Example Question #41 : Snell's Law

Light in a medium with index of refraction $n_{1}$ arrives at a boundary with another medium (with index of refraction $n_{2}$ ) at an angle of $30\degree$ to the normal. The refracted light exits the boundary at an angle of $50\degree$ from the normal. What is $\frac{n_{2}}{n_{1}}$ ?

Possible Answers:

1.53

1.347

Correct answer:

Explanation:

This is a direct application of snell's law.

$n_{1}sin(\theta _{1})=n_{2}sin(\theta _{2})$

$n_{1}sin(30\degree)=n_{2}sin(50\degree)$

$\frac{n_{2}}{n_{1}}=\frac{sin(30\degree)}{sin(50\degree)}$

This gives

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Example Question #48 : Snell's Law

How long does it take light to travel 10.0m in water ( n=1.33 )?

Possible Answers:

4.4*10^7s

$4.4*10^{-9}s$

$4.4*10^{-8}s$

$4.4*10^{-7}s$

4.4*10^8s

Correct answer:

$4.4*10^{-8}s$

Explanation:

Using Snell's law:

$v=\frac{c}{n}=\frac{3*10^8}{1.33}=225563910\frac{m}{s}$

To find the time, we need to use $v=\frac{x}{t},$ where is the distance traveled. Putting in our values, we get:

$t=\frac{x}{v}=\frac{10}{225563910}=4.4*10^{-8}s$

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Example Question #49 : Snell's Law

A laser beam traveling from air to water ( n=1.33 ) at an angle $35^{\circ}$ relative to normal. What is the angle of refraction for this scenario?

Possible Answers:

$33.1^{\circ}$

$24.2^{\circ}$

$27.9^{\circ}$

$13.1^{\circ}$

$41.8^{\circ}$

Correct answer:

$24.2^{\circ}$

Explanation:

Using Snell's law:

$n_i sin\theta_i=n_r sin\theta_r$ . The subscript stands for incident, while the stands for refraction. Here, we want to solve for the refraction angle. Let's plug in numbers:

$1sin(35\, ^{\circ})=1.33sin\theta_r$

$sin^{-1}(\frac{sin35^{\circ}}{1.33})=24.2^{\circ}$

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Example Question #50 : Snell's Law

A laser beam traveling from air to glass ( n=1.5 ) at an angle $90^{\circ}$ relative to normal. What is the angle of refraction for this scenario?

Possible Answers:

$42^{\circ}$

$30^{\circ}$

$45^{\circ}$

$24^{\circ}$

$50^{\circ}$

Correct answer:

$42^{\circ}$

Explanation:

Snell's Law states:

$n_i sin\theta_i=n_r sin\theta_r$ . The subscript stands for incident, while the stands for refraction. Here, we want to solve for the refraction angle. Let's plug in numbers:

$1sin(90^{\circ})=1.5sin\theta_r$

$sin^{-1}(\frac{1}{1.5})=42^{\circ}$

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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 #41 : Snell's Law

Example Question #42 : Snell's Law

Example Question #43 : Snell's Law

Example Question #71 : Optics

Example Question #45 : Snell's Law

Example Question #46 : Snell's Law

Example Question #41 : Snell's Law

Example Question #48 : Snell's Law

Example Question #49 : Snell's Law

Example Question #50 : Snell's Law

All AP Physics 2 Resources

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