We can use Snell's law:

At the first instance of total internal reflection, the angle of light on the air side will be  to the vertical, so the equation becomes:

Rearranging for the index of refraction of water, we get:

What this question is essentially asking for is the critical angle, which is the angle at which total internal reflection occurs. When this happens, the light will bend at an angle of  from the normal, or essentially along the interface of the two media. To solve for this value, we'll need to make use of Snell's law:

Where  is the index of refraction for benzene and  is the index of refraction for water.

To determine the critical angle, we'll make the following adjustments:

And plugging in values:

This is the angle where the incident light ray will be refracted at exactly . Any incident angle larger than this value will result in total internal reflection, where the light ray will essentially bounce back into the medium in which it originated.

Also, one very important thing to keep in mind is that total internal reflection can only occur when light travels from a medium with a higher index of refraction to a second medium with a lower index of refraction. So in the example given, total internal reflection can only occur when light travels from benzene (higher index) to water (lower index) and not the other way around.

The equation to find total internal reflection is a special case of Snell's Law, with the refraction angle set to . This new equation looks like this:

To find the critical angle, we can rearrange the equation.

As you can tell from this, the index of refraction of the medium the light is traveling from must be higher than the index of refraction of the other medium. Our numbers qualify. Now, we can plug in our values to find the critical angle.

To find the critical angle in total internal reflection problems, the following equation must be used:

 where the subscripts  stands for critical,  stands for refraction, and  stands for incident index of refraction. 

Plug in the values and solve for the critical angle:

Here, the incident index of refraction belongs to water.

For this question, we're told that light is traveling from one medium into another one. We're given the indices of refraction for both media, as well as the incident angle with respect to the normal, and we're asked to find the critical angle.

It's important to remember that when light crosses from one medium into a different one, the speed at which the light travels will change. Depending on how fast the light travels in a given medium compared to its speed in a vacuum will determine its index of refraction. In a denser medium, the light will bend towards the normal more heavily, while in a less dense medium it will bend farther from the normal.

For there to be a critical angle, light must travel from a medium of greater density to a medium of lesser density. In other words, the light must travel from a medium with a higher index of refraction into one with a lower index of refraction. Hence, there will not be a critical angle if the ray of light is traveling from a medium with a low index of refraction to a second medium with a higher index of refraction. In the question stem, this is exactly what is happening. Consequently, there cannot be a critical angle.

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

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

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

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

Simplifying, we get:

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

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

Therefore, the angle of incidence is .

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.

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

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

We expect that the index of refraction will be greater than that of air and a velocity less than .

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

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