To have total internal reflection, our equation will become:

$\displaystyle 1*sin(45^o)=n_2sin(90^o)$

$\displaystyle sin(45^o)=n_2=.71$

We can use our knowledge about the indices of refraction to come up with our equation:

$\displaystyle n_1sin(45^o)=n_2sin(22^o)$, where $\displaystyle n_1$ is the index of refraction of air and $\displaystyle n_2$ is the index of refraction for our alcohol. 

Since $\displaystyle n_1=1$

$\displaystyle \frac{sin(45^o)}{sin(22^o)}=n_2=1.9$

For this question, we're told that light is passing from air into another medium. In doing so, the light is refracted such that it bends away from the normal. We're asked to identify a possible medium.

The most important thing to understand about refraction is that when light passes into a new medium at an angle with respect to the normal, that light will be refracted, either away from or toward the normal. This is because light will travel through different media at different speeds. The faster light travels, the more it will bend away from the normal.

Generally, the angle at which light is bent can be predicted by Snell's law. In doing so, this equation takes use of the refractive index, a value unique to each medium. The expression for refractive index is as follows.

$\displaystyle n=\frac{c}{v}$

Where $\displaystyle n$ refers to the refractive index, $\displaystyle c$ refers to the speed of light in a vacuum, and $\displaystyle v$ refers to the speed of light in a given medium.

Since the speed of light is fastest when in a vacuum, the refractive index can never be less than $\displaystyle 1$. Only when the light is in a vacuum is the refractive index equal to $\displaystyle 1$. In any other medium, the refractive index will be greater than $\displaystyle 1$, even if it is slight.

Light will always refract away from the normal when it passes into a medium with a lower refractive index (indicating the light is traveling faster). Starting from air, the only way this can happen is for the new medium to have an index of refraction that is less than air. Of the answer choices shown, the only one that fits that criteria is the vacuum.

The definition of refractive index of a medium $\displaystyle n$ is the speed of light in a vacuum divided by the speed of light in the medium:

$\displaystyle n = \frac{c}{v}$

We have values for $\displaystyle c$ and $\displaystyle v$, so we can plug in our numbers into the equation.

$\displaystyle n&=\frac{2.99\cdot 10^8\ \frac{m}{s}}{1.25\cdot 10^8\ \frac{m}{s}}$

$\displaystyle n= 2.392$

Because we're dividing two values with the same units, our answer is unitless. 

In a converging lens, the light waves pass through it and have their angles altered so that they point closer together than they did before they went through the lens. In the picture, the light waves are diverging from a point until they go into the lens, at which point they no longer diverge from each other. Therefore, this is a converging lens. Because the waves are travelling the same direction the whole time, it can't be the converging or diverging mirrors. If the lens were diverging, they'd be more separated. If it were a plane mirror, the waves would get polarized (they'd have the same phase angle).

Use the formula for focal length.

$\displaystyle M=\frac{\frac{h_0}{f}}{\frac{h_0}{N}}$

Solve for focal length by substituting known values.

$\displaystyle f=\frac{N}{M} = \frac{20 { {cm}}}{5}=4 { cm}=0.04{ m}$

Using Malus' law.

$\displaystyle I=I_0cos^2(\theta)$

Since the initial light is unpolarized, there will be no intensity lost.

$\displaystyle I=I_0cos^2(25^{\circ})$

$\displaystyle I=I_0*.821$

$\displaystyle 82.1\%=.821$

Use Malus' law.

$\displaystyle I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. It is thus necessary to use Malus' law twice.

$\displaystyle I_1=I_0cos^2(\theta_1)$

$\displaystyle I_2=I_1cos^2(\theta_2)$

Where $\displaystyle I_0$ is the initial intensity after the first polarizer.

$\displaystyle I_1$ is the intensity after the second polarizer.

$\displaystyle I_2$ is the intensity after the third polarizer.

$\displaystyle \theta_1$ is the angle between the first and second polarizers.

$\displaystyle \theta_2$ is the angle between the second and third polarizers.

Combining equations:

$\displaystyle I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$\displaystyle I_2=I_0cos^2(25^\circ)cos^2(35^\circ)$

$\displaystyle I_2=.551I_0$

Use Malus' law.

$\displaystyle I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus' law twice.

$\displaystyle I_1=I_0cos^2(\theta_1)$

$\displaystyle I_2=I_1cos^2(\theta_2)$

Where $\displaystyle I_0$ is the initial intensity after the first polarizer.

$\displaystyle I_1$ is the intensity after the second polarizer.

$\displaystyle I_2$ is the intensity after the third polarizer.

$\displaystyle \theta_1$ is the angle between the first and second polarizers.

$\displaystyle \theta_2$ is the angle between the second and third polarizers.

Combining equations:

$\displaystyle I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$\displaystyle I_2=I_0cos^2(65^\circ)cos^2(55^\circ)$

$\displaystyle I_2=.0588I_0$

Use Malus' law.

$\displaystyle I=I_0cos^2(\theta)$

The light's intensity is reduced by the final two polarizers. We will need to use Malus' law twice.

$\displaystyle I_1=I_0cos^2(\theta_1)$

$\displaystyle I_2=I_1cos^2(\theta_2)$

Where $\displaystyle I_0$ is the initial intensity after the first polarizer.

$\displaystyle I_1$ is the intensity after the second polarizer.

$\displaystyle I_2$ is the intensity after the third polarizer.

$\displaystyle \theta_1$ is the angle between the first and second polarizers.

$\displaystyle \theta_2$ is the angle between the second and third polarizers.

Combine equations:

$\displaystyle I_2=I_0cos^2(\theta_2)cos^2({\theta_1})$

Plug in values:

$\displaystyle I_2=I_0cos^2(45^\circ)cos^2(5^\circ)$

$\displaystyle I_2=.496I_0$