This question asks us to consider the relationship between velocity and index of refraction of a medium. If we think back to the definition of index of refraction, we know that it is defined by the ratio of the velocity of light in a vacuum and the velocity of light in some other medium.

n is the index of refraction, c is the speed of light in a vacuum, and v is the speed of light in the new medium.

We can see that n and v are inversely proportional, meaning that the higher the n, the lower the velocity. As the light moved from air (n =1) to glass (n = 1.5), the n increased, and thus the velocity must decrease because the speed of light in a vacuum is constant.

This question asks us to find the relationship between the wavelength and index of refraction. We will need to know two equations to compute the relationship between the two.

First, we need to relate velocity and index of refraction. The definition of index of refraction allows us to relate the two.

We also know the relationship between velocity and wavelength.

 We can now set these formulas equal to each other to find the relationship between wavelength and index of refraction.

We can see that  and n are inversely related. If n decreases, the wavelength must increase. In our problem, light in moving from a higher index of refraction to a lower one, meaning the wavelength gets longer (increases).

The index of refraction is equal to the speed of light in a medium divided by the speed of light in a vacuum.

We can find the time to travel a given distance by manipulating this equation and combining it with the equation for rate: .

Plug in the given values and solve for the velocity in the medium.

Now we can return to the rate equation and solve for the time to travel .

We can recognize that the answer can be simplified by converting to nanoseconds.

The refractive index of a medium (n) is equal to the speed of light (c) divided by the velocity of light through the medium (v).

Rearranging the equation allows us to see the relationship regarding v.
 
The lower the refractive index, the faster the velocity of light. Medium A has the smaller refractive index. Light will travel faster through medium A at a velocity equal to the speed of light divided by the refractive index.

As the wave travels into the less dense medium, it speeds up, bending away from the normal line. The index of refraction tells the ratio of the velocity in a vacuum in relation to the velocity the medium; thus, the velocity will be greater in a medium with a lower index of refraction.

Frequency remains the same regardless of medium, however, since the velocity changes, the wavelength must accommodate this change.

If velocity increases and frequency remains constant, wavelength must also increase.

Finally, a phase shift only occurs when a light ray reflects from the surface of a more dense medium.

Relevant equations:

To find index of refraction, divide the speed of light in a vacuum by the speed of light in the material:

To compare angles of incidence and refraction, use Snell's law.

Notice that as the light enters a more dense medium, it bends towards the normal.

This problem asks us to consider refraction, that is, the bending of light when it enters a new medium. Thinking back to our light formulas in physics, we know that , where n is the index of refraction of the medium and  is the angle the light ray makes to normal.

Using the information provided in the pre-question text and the question above, we can solve for .

Notice that, in this problem, there are two different changes in indices of refraction: air to glass and glass to solution; thus, we need to compute the angle change twice. First, let’s look at how the angle changes from air to glass. We know from Snell’s law that , where n is the index of refraction of the medium and  is the angle the light ray makes to normal.

Rearranging, we can find the angle to normal in the glass.

Now, we can use this angle and repeat the above equation to find the angle that the light enters the solution.

This question is asking us to consider what would happen if the solution were more concentrated, essentially, if there were more particles per unit volume (denser). If we think back to the definition of index of refraction, we know that it relates to the density of a medium. The denser the medium, the higher the index of refraction.

Looking at Snell’s law, we can see the relationship between index of refraction and the angle of refraction.

If the solution were more concentrated, n₂ would increase, making the term on the right side of the equation smaller. The sin function of a smaller number gives a smaller angle; thus, as the concentration increases (and thus the index of refraction increases), the angle of refraction gets smaller.