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AP Physics 2 : Snell's Law

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Example Question #1 : Snell's Law

A ray of light is traveling through air at an angle of $30^{\circ}$ to the vertical. It passes into water and halves its angle to the vertical. What is the index of refraction of water?

$\theta_{air}=1.00$

Possible Answers:

1.00

1.93

1.54

0.88

0.52

Correct answer:

1.93

Explanation:

This problem can be solved using Snell's law:

$n_1sin(\theta_1) = n_2sin(\theta_2)$

Rearrange to solve for for the index of refraction of water

$n_2 = \frac{n_1sin(\theta_1)}{sin(\theta_2)}$

We can determine the second angle:

$\theta_2 = \frac{\theta_1}{2} = 15^{\circ}$

We have all the values for Snell's law, allowing us to solve:

$n_2 = \frac{(1.0)sin(30)^{\circ}}{sin(15^{\circ})} = 1.93$

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Example Question #1 : Snell's Law

A beam of light going through a medium with an index of refraction of 1.14 has an angle of incidence of $48^{\circ}$ with another medium. If the light wave is at a new angle of $39^{\circ}$ , what is the index of refraction for the second medium?

Possible Answers:

1.55

There is not enough information to determine the index of refraction.

1.346

0.98

1.14

Correct answer:

1.346

Explanation:

To find the index of refraction, we use Snell's Law:

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

We have values for n_1 _,, and $\theta_2$ , so now we just need to rearrange the equation to solve for n_2 .

$\begin{align*} n_1\sin\theta_1&=n_2\sin\theta_2 \\ n_1\frac{\sin\theta_1}{\sin\theta_2}&=n_2 \end{align*}$

Now, we can just plug in our numbers.

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

$n_2=(1.14)(\frac{\sin(48^o)}{\sin(39^o)})$

n_2= 1.346

Therefore, the index of refraction of the second material is 1.346.

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Example Question #3 : Snell's Law

Find the angle of refraction if a light beam enters water at an angle of $10^\textup{o}$ relative to the normal.

$n_{water}=1.33$ and $n_{air} = 1$

Possible Answers:

$7.5^{\textup o}$

$15^{\textup o}$

$1.125^{\textup o}$

$10.5^{\textup o}$

$3.25^{\textup o}$

Correct answer:

$7.5^{\textup o}$

Explanation:

To find the angle of refraction, use the Snell's law equation to determine the angle.

$n_1sin(\theta_{1})=n_2sin(\theta_2)$

Find the index of refraction for water.

$n_2=n_{water}=1.33$

Find the index of refraction for air.

$n_1=n_{air}=1$

Solve for $\theta_2$ .

$(1)sin(10^{\textup o})=(1.33)sin(\theta_2)$

$\theta_2=sin^{-1}({\frac{sin(10^{\textup o})}{1.33}})=7.5^{\textup o}$

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Example Question #1 : Snell's Law

A cup of water sitting on a table outside in the sunlight is filled to the top. The sun happens to be $52^{o}$ above the horizon and you notice the bottom of the cup has just become completely shaded. The cup has a depth of 20cm and you realize you can calculate the width of the bottom of the cup. Find the width in .

Possible Answers:

w=20.1cm

w=2.5cm

w=12.43cm

w=10.44cm

Correct answer:

w=10.44cm

Explanation:

This problem is a Snell's Law problem. There is some information not stated but can be found in your text book such as the index of refraction of water. A picture will help us see what is going on in this problem,

Cup

$\alpha$ will be defined as the angle w.r.t. the horizon that the sun is at. Let be the depth of the water and since the light passes through air $\left ( n_{1}=1\right )$ into water $\left ( n_{2}=1.33\right )$ , it is just an application of Snell's Law, then a simple trig calculation.

First, let's find $\theta_{1}$ ,

$\theta_{1}=90^{o}-\alpha=38^{o}$

Now apply Snell's Law to find $\theta_{2}$ ,

$n_{1}\sin\left ( \theta_{1}\right )=n_{2}\sin\left ( \theta_{2}\right )$

$\theta_{2}=\arcsin \left [ \frac{n_{1}}{n_{2}}\sin\left ( \theta_{1}\right ) \right ]=27.57^{o}$

Now there is a right triangle with sides and and an angle of $\theta_{2}$ . Using the tangent,

$\tan\left (\theta_{2}\right )=\frac{w}{d} \rightarrow w=d \tan\left (\theta_{2}\right )=10.44cm$

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Example Question #1 : Snell's Law

A ray of light initially travelling through the air enters a material at a angle of $30^{\circ}$ with respect to the line normal to the surface. The light is this observed to bend toward the normal by $5^{\circ}$ . Calculate the index of refraction ,, of the material.

Possible Answers:

1.20

0.55

2.05

1.0

1.18

Correct answer:

1.18

Explanation:

By Snell's law, we have the relation between the incident angle $\theta _1$ , the refracted angle $\theta _2$ , and the indices of refraction of the two materials n_1 and n_2 given by:

$n_1 \sin \theta _1 = n_2 \sin \theta _2$

The index of refraction of air is taken to be $1.003 \approx 1.0$ . By substituting in the given values into our rearranged equation, we can solve for the index of refraction of the material.

$n_2 = n_1 \frac{\sin \theta _1}{\sin \theta _2}$

$1.0\frac{\sin 30^\circ}{\sin 25^{\circ}} = 1.18$

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Example Question #1 : Snell's Law

A laser initially moving through air enters a medium that has an index of refraction n_2=1.40 . The beam's angle with respect to the normal inside the material is measured to be $\theta _2 = 32^{\circ}$ . Calculate the incident angle (the angle the laser made with respect to the normal before entering the material).

Possible Answers:

$1.0^{\circ}$

$51.6^{\circ}$

$47.9^{\circ}$

$-14.3^{\circ}$

$23.9^{\circ}$

Correct answer:

$47.9^{\circ}$

Explanation:

Using Snell's Law, we can use the information given to solve for $\theta _1$ . This is shown below as:

$n_1 \sin \theta _1 = n_2 \sin \theta _2$

Solve for the incident angle.

$\theta _1 = \sin ^{-1} \left(n_2 \sin \theta _2 \right )$

Plug in and solve:

$\theta_1= \sin ^{-1} \left(1.40 \sin 32^{\circ} \right )=47.9^{\circ}$

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Example Question #7 : Snell's Law

Suppose that a ray of light traveling through air strikes the surface of a body of water, and the ray of light then begins traveling through the water. If the ray of light from air hits the surface of the water at an angle of $60^{o}$ from the horizontal, then how far from the horizontal will the ray of light in water be bent?

$n_{air}=1$

$n_{water}=1.33$

Possible Answers:

$22.08^{o}$

$40.62^{o}$

$67.92^{o}$

$49.38^{o}$

Correct answer:

$67.92^{o}$

Explanation:

This question is presenting us with a scenario in which light is first traveling through air, and then it hits the surface of water and begins traveling through the water. We're provided with the angle of the incident light, and are asked to find the angle of the light with respect to the horizontal after refraction has occurred.

To answer this question, we'll need to make use of Snell's law:

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

Where $n_{1}$ and $n_{2}$ represent the index of refraction for air and water, respectively. Furthermore, $\Theta _{1}$ is the angle of the incident light ray (from air) with respect to the normal, while $\Theta _{2}$ is the angle of the refracted ray in water with respect to the normal.

When establishing the values for the variables listed above, it's important to be aware of the difference between the angle with respect to the horizontal (as presented in the question stem) and the angle with respect to the normal (which is perpendicular to the surface boundary). In the question stem, we're told that the incident light ray strikes the surface boundary at an angle of $60^{o}$ with respect to the horizontal. This means that the angle with respect to the normal will be $30^{o}$ .

Now, we have all the information we need to solve for the missing angle. But first, we can rearrange the above equation to isolate the variable we're looking for.

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

$\Theta_{2}=\sin^{-1}(\frac{1}{1.33}* sin30^{o})$

$\Theta_{2}=\sin^{-1}(\frac{1}{1.33}* 0.5)$

$\Theta_{2}=\sin^{-1}(0.376)$

$\Theta_{2}=22.08^{o}$

But there's one more very important thing we have to do. Recall that the question is asking for the resulting angle with respect to the horizontal, whereas this answer is telling us the angle with respect to the normal. Therefore, we will need to convert this angle:

$90^{o}-22.08^{o}=67.92^{o}$

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Example Question #8 : Snell's Law

Light travels from air n=1 to water n=1.33 . If the angle of incidence is $45^\circ$ , determine the angle of refraction.

Possible Answers:

$63.3^\circ$

$17.7^\circ$

$52.5^\circ$

$32.1^\circ$

None of these

Correct answer:

$32.1^\circ$

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

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:

$sin^{-1}(\frac{1*sin45^\circ}{1.33})=\theta_2$

Solving:

$32.1^\circ=\theta_2$

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Example Question #1 : Snell's Law

Light travels from air n=1 to water n=1.33 . If the angle of incidence is $27^\circ$ , determine the angle of refraction.

Possible Answers:

$29.9^\circ$

$20.0^\circ$

None of these

$25.5^\circ$

$21.6^\circ$

Correct answer:

$20.0^\circ$

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

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:

$sin^{-1}(\frac{1*sin27^\circ}{1.33})=\theta_2$

Solve:

$20.0^\circ=\theta_2$

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Example Question #10 : Snell's Law

Light travels from air n=1 to water n=1.33 . If the angle of incidence is $15^\circ$ , determine the angle of refraction.

Possible Answers:

None of these

$8.2^\circ$

$22.5^\circ$

$17.5^\circ$

$11.2^\circ$

Correct answer:

$11.2^\circ$

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

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:

$sin^{-1}(\frac{1*sin15^\circ}{1.33})=\theta_2$

$11.2^\circ=\theta_2$

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AP Physics 2 : Snell's Law

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