Snell's Law:

\(\displaystyle \frac{sin\theta_1}{sin\theta_2}=\frac{v_1}{v_2}=\frac{\lambda_1}{\lambda_2}=\frac{n_2}{n_1}\)

Where

\(\displaystyle \theta\) is the respective incident angle

\(\displaystyle v\) is the respective velocity of the light

\(\displaystyle \lambda\) is the respective wavelength

\(\displaystyle n\) is the respective index of refraction

Use the following version of Snell's law:

\(\displaystyle \frac{sin\theta_1}{sin\theta_2}=\frac{n_2}{n_1}\)

Solving for \(\displaystyle n_2\)

\(\displaystyle n_1\frac{sin\theta_1}{sin\theta_2}=n_2\)

Plugging in values

\(\displaystyle 1*\frac{sin77^\circ}{sin68^\circ}=n_2\)

\(\displaystyle 1.05=n_2\)

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.

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

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

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

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

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

Using Snell's Law:

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

The index of refraction of a vacuum is unity, \(\displaystyle n=1\)

Plugging in values:

\(\displaystyle 1sin45^\circ=n_2sin18^\circ\)

\(\displaystyle n_2=2.29\)

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

\(\displaystyle \frac{c}{\frac{2d}{t}}\)

Which when simplified, gives:

\(\displaystyle \frac{ct}{2d}\)

Using Snell's law:

\(\displaystyle \frac{n_1}{n_2}=\frac{v_2}{v_1}=\frac{sin\theta_2}{sin\theta_1}=\frac{\lambda_2}{\lambda_1}\)

\(\displaystyle \frac{1}{n_2}=\frac{sin15^\circ}{sin33^\circ}\)

\(\displaystyle n_2=2.10\)

Using Snell's law:

\(\displaystyle \frac{n_1}{n_2}=\frac{v_2}{v_1}=\frac{sin\theta_2}{sin\theta_1}=\frac{\lambda_2}{\lambda_1}\)

\(\displaystyle \frac{sin 15^\circ}{sin 33^\circ}=\frac{\lambda_2}{650nm}\)

\(\displaystyle \lambda_2=309nm\)

This is a direct application of snell's law. 

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

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

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

This gives

\(\displaystyle .653\)

Using Snell's law:

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

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

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

Using Snell's law:

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

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

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

Snell's Law states:

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

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

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