Snell's Law - AP Physics 2

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Question

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?

$\theta_{air}=1.00$

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 $48^{\circ}$ 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?

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.

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

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 $\theta_2$ .

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

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

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Question

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 and you realize you can calculate the width of the bottom of the cup. Find the width in .

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 $\theta_{1}$ ,

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

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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Question

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

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 $\theta_2$ :

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

Plug in values:

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

Solving:

$32.1^\circ=\theta_2$

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Question

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

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 $\theta_2$ :

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

Plug in values:

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

Solve:

$20.0^\circ=\theta_2$

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Question

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

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 $\theta_2$ :

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

Plug in values:

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

$11.2^\circ=\theta_2$

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Question

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

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 $\theta_2$ :

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

Plug in values:

$sin^{-1}(\frac{1*sin45^\circ}{1.52})=\theta_2$

$27.7^\circ=\theta_2$

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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.

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 $\theta_2$ :

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

Plug in values:

$sin^{-1}(\frac{1*sin22^\circ}{1.52})=\theta_2$

$14.3^\circ=\theta_2$

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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.

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 $\theta_2$ :

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

Plug in values:

$sin^{-1}(\frac{1*sin78^\circ}{1.52})=\theta_2$

$40.1^\circ=\theta_2$

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Question

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

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.

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.

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.

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.

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.

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?

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?

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:

$v_1=c=3.010^8\frac{m}{s}$

Plug in values:

$\frac{1}{1.24}3.010^8\frac{m}{s}=v_2$

$2.41*10^8\frac{m}{s}=v_2$

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Question

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

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:

$v_1=c=3.010^8\frac{m}{s}$

Plug in values:

$\frac{1}{1.74}3.010^8\frac{m}{s}=v_2$

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

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Question

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

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,

$v_1=c=3.010^8\frac{m}{s}$

Plug in values:

$\frac{1}{1.56}3.010^8\frac{m}{s}=v_2$

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

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