1. Complete laser mini-labs:

3. Solve Problem #3 or #4 (below). You may wish to refer to the formulation (eq.22.2) of Snell's Law on p.691 (Wilson & Buffa, 4th Ed.):
n = c/v so  n₁sinq₁ = n₂sinq₂ = csinq₁/v₁ = csinq₂/v₂

1. Blue light of wavelength 470 nm strikes the side of an equilateral prism.(each angle is 60^o). The angle of incidence is 53.8^o. The index of refraction for blue light is 1.636 in this prism.
a. Calculate the refracted angle relative to a line perpendicular to the first air-glass interface.
b. If the prism has three 60^o angles, calculate the incident angle at the second glass-air interface.
c. Calculate the refracted angle relative to a line perpendicular to this surface.
[Ans: a. 29.6^o b. 30.4^o c. 56^o]

2. Repeat problem #1, but imagine that the prism is immersed in water. Assume that the index of refraction of water for blue light of this wave length is 1.330.

3. Sound passes from limestone, where it moves at a speed of 4000 m/s, into another unknown material. The angle of incidence at the interface is 24^o and the angle of refraction in the unknown material is 38^o.
a. Calculate the speed of sound in the unknown material.
b. Make a qualitative sketch of the situation and identify relevant quantities.
[Ans: a. 6050 m/s]

4. Sound traveling in limestone enters the top surface of a cubical salt dome at an angle of incidence of 35^o. After passing through the salt dome, the sound leaves the right side and reenters the limestone. Calculate its final direction in the limestone and show in a figure the directions of the ray as it enters, passes through, and leaves the salt dome.
(v_limestone = 4000 m/s and v_salt = 6000 m/s) {Ans: 20^o below horizontal}

Background:

In any homogeneous material, light travels in straight lines. However, when light encounters a boundary, a change in material, some of the light reflects back and some of the light is transmitted forward into the new material. The transmitted light does not travel in the same direction as the original light. Instead it is bent at the boundary and travels in a slightly different direction. When light is bent in this way, the phenomenon is called refraction.

The refraction of light at the boundary between two materials is described quantitatively by Snell's Law. Consider the situation depicted in Figure 1. A ray of light is incident on the surface between two different materials. The long dashed line represents a line perpendicular to the surface. The angle q measures the angle between the incident ray and the line perpendicular to the surface. This angle is the angle of incidence. The angle f measures the angle between the refracted ray and the line perpendicular to the surface. This angle is the angle of refraction. Snell's Law states:

The quantity n_i is the index of refraction for the material in which the light was incident. The quantity n_r is the index of refraction for the material in which the light was refracted.

The phenomenon of total internal reflection can occur when the light travels from a material with higher index of refraction to a material with a lower index of refraction. When n_i > n_r, the refracted ray is bent away from the line perpendicular to the surface. If the angle of incidence is large enough, the angle of refraction will be 90^o, and the light is refracted parallel to the surface. The angle of incidence for which this occurs is called the critical angle. If the angle of incidence is any greater than the critical angle, it is not possible for there to be a refracted ray. [Applying Snell's Law for this case would lead one to predict that sin(f) > 1, which contradicts what is known of the sine function ( -1< sin(f) < 1).] When there is no refracted ray, all of the energy of the light goes into the reflected ray.

Through investigation of total internal reflection one can determine the index of refraction of various substances. By measuring the critical angle for a situation in which light goes from a material of higher index of refraction to a material of lower index of refraction, and by knowing the index of refraction of one of the materials, you can determine the index of refraction of the other material. The critical angle q_crit is the largest angle for which there is not evidence of total internal reflection. The angle of refraction would be 90^o in that case. Therefore, sin(f) = 1, because f = 90^o.  Air has an index of refraction equal to 1. If the second material, into which the light was traveling, was air, then n_r = 1. If this information is substituted into Snell's Law:

The phenomenon of total internal reflection is important in numerous technologies. Total internal reflection explains how light can be shined into one end of a glass fiber and transmitted great distances with little loss of energy. Such fiber optics are used in electronic communication to replace copper wires and in modern diagnostic and surgical medical devices, which minimize the amount of trauma to the body.

Materials:
semi-circular plastic dish,   protractor or polar graph paper
laser pen pointer,     non-dairy creamer
water,  aqueous cane sugar solution (i.e. sugar-water)