The angle of reflection is the angle between the reflected light ray and a line perpendicular to the interface between the two media. The angle of reflection must be complementary to 20^o.

90^o – 20^o = 70⁰

This question is asking us about the different processes that can happen to light upon hitting a surface. Thinking back to the properties of light, we know that absorption, reflection, and refraction are all processes that light can undergo when interacting with a surface.

Absorption means that the energy associated with the light is captured, and no photons are ejected from the surface after the collision of the incident photon and the surface. Reflection occurs when no light enters the new medium and instead bounces off at the angle to normal that it hit. Refraction occurs when some light enters the new medium. In this case, light is entering glass from the air; thus, the process we are concerned about is refraction. 

For all waves, the angle of incidence is equal to the angle of reflection. The projection is to the opposite side of the normal at the same angle as the incident wave.

Think of the wall as a mirror. The angle with which the wave impacts a mirror will be equal to the angle with which it is reflected, but mirrored across the normal.

Since the index of refraction is 1.5, we can determine the speed of light in the glass using the following equation:

Rearranged to solve for velocity:

Once the speed of light in the glass is known, we can use this quantity to determine how long it will take for the light to travel the width of the glass .

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: