AP Physics 2 - Snell's Law

Light travels from air to diamond . If the angle of incidence is $75^\circ$ , determine the angle of refraction.

$23.5^\circ$

None of these

$47.5^\circ$

$56.0^\circ$

$18.9^\circ$

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

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:

$sin^{-1}(\frac{1*sin75^\circ}{2.42})=\theta_2$

$23.5^\circ=\theta_2$

Light in a vacuum hits the surface of an unknown material with an angle of incidence of $29^\circ$ . The angle of refraction is $10^\circ$ . What is the index of refraction of the unknown material?

None of these

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 :

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

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

Plugging in values

$1*\frac{sin29^\circ}{sin10^\circ}=n_2$

Light of wavelength 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.

None of these

Explanation

Using Snell's law:

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

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

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?

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

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

Plugging in values

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

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

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

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

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

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

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

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 :

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

The velocity of light in a vacuum is

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

Plug in values:

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

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

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

$2.04*10^8\frac{m}{s}$

$2.65*10^8\frac{m}{s}$

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

None of these

$2.11*10^8\frac{m}{s}$

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 :

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

The velocity of light in a vacuum:

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

Plug in values:

$\frac{1}{1.24}3.0*10^8\frac{m}{s}=v_2$

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

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 $\frac{n_{2}}{n_{1}}$ ?

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

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?

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

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 , $\theta_1$ , 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.

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

$42^{\circ}$

$45^{\circ}$

$50^{\circ}$

$30^{\circ}$

$24^{\circ}$

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:

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

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

Light of wavelength 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.