Snell's Law - AP Physics 2

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Question

A ray of light is traveling through air at an angle of to the vertical. It passes into water and halves its angle to the vertical. What is the index of refraction of water?

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Answer

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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Question

A beam of light going through a medium with an index of refraction of 1.14 has an angle of incidence of with another medium. If the light wave is at a new angle of , what is the index of refraction for the second medium?

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Answer

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 , $\theta_1$ , and $\theta_2$ , so now we just need to rearrange the equation to solve for .

$\begin{align} n_1\sin\theta_1&=n_2\sin\theta_2 \ n_1$\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$)}$)$

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

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Question

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

and

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Answer

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 $\theta_2$ .

$\theta_$2=sin^{-1}$$({$\frac{sin(10^{\textup o}$$$)}{1.33}})=7.5^{\textup o}$$

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Question

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

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Answer

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,

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

Now apply Snell's Law to find ,

Now there is a right triangle with sides and and an angle of . 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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Question

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

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Answer

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

Use Snell's Law

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

Solve for $\theta_2$ :

Plug in values:

Solving:

$32.1^\circ=\theta_2$

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Question

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

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Answer

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

Use Snell's Law

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

Solve for $\theta_2$ :

Plug in values:

Solve:

$20.0^\circ=\theta_2$

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Question

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

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Answer

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

Use Snell's Law

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

Solve for $\theta_2$ :

Plug in values:

$11.2^\circ=\theta_2$

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Question

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

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Answer

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

Use Snell's law

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

Solve for $\theta_2$ :

Plug in values:

$27.7^\circ=\theta_2$

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Question

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

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Answer

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

Use Snell's law

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

Solve for $\theta_2$ :

Plug in values:

$14.3^\circ=\theta_2$

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Question

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

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Answer

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

Use

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

Solve for $\theta_2$ :

Plug in values:

$40.1^\circ=\theta_2$

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Question

Light travels from air to water . If the wavelength of incident light is , determine the wavelength of refracted light.

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Answer

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

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}$530nm$

$\lambda_2=398nm$

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Question

Light travels from air to water . If the wavelength of incident light is , determine the wavelength of refracted light.

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Answer

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

Use

$\frac{\lambda_1}{\lambda_2}$=\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}$470nm$

$\lambda_2=353nm$

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Question

Light travels from air to water . If the wavelength of incident light is , determine the wavelength of refracted light.

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Answer

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

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}$610nm$

$\lambda_2=459nm$

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Question

Light travels from air to crown glass . If the wavelength of incident light is , determine the wavelength of refracted light.

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Answer

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

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}$530nm$

$\lambda_2=349nm$

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Question

Light travels from air to crown glass . If the wavelength of incident light is , determine the wavelength of refracted light.

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Answer

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

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}$620nm$

$\lambda_2=408nm$

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Question

Light travels from air to crown glass . If the wavelength of incident light is , determine the wavelength of refracted light.

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Answer

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

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}$690nm$

$\lambda_2=454nm$

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Question

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?

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Answer

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

Use Snell's Law:

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

Solve for :

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

Plug in values:

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

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Question

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

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Answer

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

Use Snell's law

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

Solve for :

$$\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:

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Question

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

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Answer

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

Use Snell's law:

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

Solve for

$$\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:

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Question

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

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Answer

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

Use Snell's law:

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

Solve for

$$\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:

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