32.1. Identify: Since the speed is constant, distance

Set Up: The speed of light is .

Execute: (a)

(b)

Evaluate: The speed of light is very great. The distance between stars is very large compared to terrestrial distances.

32.3.Identify: Apply

Set Up:

Execute: (a)

(b)

(c)

(d)

Evaluate: f increases when decreases.

32.5.Identify: . . .

Set Up: Since the wave is traveling in empty space, its wave speed is .

Execute: (a)

(b)

(c) . .

Evaluate: The factor is common to both the electric and magnetic field expressions, since these two fields are in phase.

32.7.Identify and Set Up: The equations are of the form of Eqs.(32.17), with x replaced by z. is along the y-axis; deduce the direction of

Execute:

is along the y-axis. is in the direction of propagation (the +z-direction). From this we can deduce the direction of as shown in Figure 32.7.

/ is along the x-axis.
Figure 32.7

Evaluate: are perpendicular and oscillate in phase.

32.9.Identify and Set Up: Compare the given in the problem to the general form given by Eq.(32.17). Use the direction of propagation and of to find the direction of

(a) Execute: The equation for the electric field contains the factor so the wave is traveling in the +y-direction. The equation for is in terms of rather than the wave is shifted in phase by relative to one with a factor.

(b)

Comparing to Eq.(32.17) gives

(c)

/ must be in the +y-direction (the direction in which the wave is traveling). When is in the –z-direction then must be in the –x-direction, as shown in
Figure 32.9.
Figure 32.9

Then

Using Eq.(32.17) and the fact that is in the direction when is in the direction,

Evaluate: and are perpendicular and oscillate in phase.

32.12.Identify:

Set Up: The magnetic field of the earth is about

Execute:

Evaluate: The field is much smaller than the earth's field.

33.25.Identify: When unpolarized light passes through a polarizer the intensity is reduced by a factor of and the transmitted light is polarized along the axis of the polarizer.When polarized light of intensity is incident on a polarizer, the transmitted intensity is , where is the angle between the polarization direction of the incident light and the axis of the filter.

Set Up: For the second polarizer .For the third polarizer, .

Execute: (a) At point A the intensity is and the light is polarized along the vertical direction.At point B the intensity is , and the light is polarized along the axis of the second polarizer.At

point C the intensity is .

(b) Now for the last filter and .

Evaluate: Adding the middle filter increases the transmitted intensity.

33.27.IdentifyandSet Up: Reflected beam completely linearly polarized implies that the angle of incidence equals the polarizing angle, so Use Eq.(33.8) to calculate the refractive index of the glass. Then use Snell’s law to calculate the angle of refraction.

Execute: (a)

(b)

Evaluate:

/ Note:Thus the reflected ray and the refracted rayare perpendicular to each other. Thisagrees with Fig.33.28.
Figure 33.27

33.29.Identify: From Malus’s law, the intensity of the emerging light is proportional to the square of the cosine of the angle between the polarizing axes of the two filters.

Set Up: If the angle between the two axes is , the intensity of the emerging light is I = Imax cos2.

Execute: At angle ,I = Imaxcos2, and at the new angle , I = Imaxcos2. Taking the ratio of the intensities gives , which gives us . Solving for  yields .

Evaluate: Careful! This result is not cos2.

33.31.Identify: When unpolarized light of intensity is incident on a polarizing filter, the transmitted light has intensity and is polarizedalong the filter axis.When polarized light of intensity is incident on a polarizing filter the transmitted light has intensity .

Set Up: For the second filter, .

Execute: After the first filter the intensity is and the light is polarized along the axis of the first filter.The intensity after the second filter is , where and .This gives