Write the formula relating the speed of light, velocity of light in specific medium, and index of refraction of the medium.

\(\displaystyle n_{material}=\frac{c}{v}\)

Rewrite the equation so that we are solving for the velocity of light in water. Substitute given values.

\(\displaystyle v=\frac{c}{n_{water}}=\frac{3 * 10^8 \frac{m}{s}}{1.33}=2.26 * 10^8\frac{m}{s}\)

To find the index of refraction given two angles and another index of refraction, we use Snell's law.

\(\displaystyle \begin{align*} n_1\sin\theta_1&=n_2\sin\theta_2 \\ \frac{n_1\sin\theta_1}{\sin\theta_2}&= n_2 \end{align*}\)

We can plug in the numbers we have to find the answer.

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

\(\displaystyle \frac{(1)(\sin(34^o))}{\sin(26^o)}&=n_2\)

Simplifying, we get:

\(\displaystyle n_2 = 1.276\)

To find the angle of incidence, we use Snell's law.

\(\displaystyle n_1\sin\theta_1&=n_2\sin\theta_2\)

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

We're given both indices of refraction, and the second angle, so we can plug in our numbers.

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

\(\displaystyle \sin \theta_1=\frac{(1.51)(\sin(41^o))}{1}\)

\(\displaystyle \theta_1 = 82.16^{\text o}\)

Therefore, the angle of incidence is \(\displaystyle 82.16^{\textup o}\).

First off we can find the velocity of the laser beam since we know the distance of honey it travels through and the transit time.

\(\displaystyle v=\frac{d}{t}\)

However, this doesn't need to be explicitly solved for. The velocity of light inside a material is inversely proportional to the index of refraction of that material where

\(\displaystyle v_{honey}=\frac{c}{n_{honey}}\)

where \(\displaystyle c\) is the speed of light in vacuum. Setting these equal, we can solve for the index of refraction of the honey.

\(\displaystyle \frac{d}{t}=\frac{c}{n_{honey}}\)

\(\displaystyle n_{honey}=\frac{c t}{d}=1.47\)

We expect that the index of refraction will be greater than that of air and a velocity less than \(\displaystyle c\).

The speed of light passing through a medium is given by:

\(\displaystyle v_{light,n}=\frac{c}{n}\)

Rearrange to solve for the index of refraction and plug in numbers.

\(\displaystyle n=\frac{c}{v_{light,n}}=\frac{3.0\times 10^{8}\frac{m}{s}}{2.5\times 10^{8}\frac{m}{s}}=1.2\)

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.