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Physics

1st Edition

Walker

ISBN: 9780133256925

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Page 553: Lesson Check

Exercise 34

Step 1

1 of 2

The direction of polarization is decided by the electric field. Hence the direction of polarization is vertical.

Result

2 of 2

Vertical.

Exercise 35

Step 1

1 of 2

The direction of the polarization of the transmitted light will be $30text{textdegree}$ from the vertical.

Result

2 of 2

$30text{textdegree}$ from vertical.

Exercise 36

Step 1

1 of 2

The key factor for the blue sky is wavelength.

Result

2 of 2

Wavelength.

Exercise 37

Step 1

1 of 2

The reflected light from the water has mostly horizontally polarized light. And the direction of polarization of sunglass is vertical, therefore when standing up or sitting upright not a lot of light is transmitted. When laying on the side the angle between the sunglasses polarizer and the incoming polarized light is reduced and more light is transmitted. Glare is less reduced.

Result

2 of 2

Glare is less reduced.

Exercise 38

Step 1

1 of 2

For $45text{textdegree}$ angle setting the angle difference from the vertical and horizontal polarization are same. So the transmitted angle will have same intensity.

Result

2 of 2

Same.

Exercise 39

Step 1

1 of 2

In the first and second case the transmitted light have same intensity. In the third case the first and second polarizer are in $90text{textdegree}$ angle hence the transmitted intensity will be zero. So in the order of increasing intensity the we have

$$
C<A=B
$$

Result

2 of 2

$$
C<A=B
$$

Exercise 40

Step 1

1 of 2

In the case A the the transmitted intensity is

$$
I_{f}=I_{i}costheta_{1}costheta_{2}=left(37.0 {rm W/m^{2}}right)left(cos45text{textdegree}right)left(cos45text{textdegree}right)=9.25
$$

In the case B, we have

$$
I_{f}=I_{i}costheta_{1}costheta_{2}=left(37.0 {rm W/m^{2}}right)left(cos45text{textdegree}right)left(cos45text{textdegree}right)=9.25
$$

In case C, we have

$$
I_{f}=I_{i}costheta_{1}costheta_{2}=left(37.0 {rm W/m^{2}}right)left(cos45text{textdegree}right)left(cos90text{textdegree}right)=0
$$

Result

2 of 2

$$
text{a)} I_{f} = 9.25
$$

$$
text{b)} I_{f} = 9.25
$$

$$
text{c)} I_{f} = 0
$$

Exercise 41

Step 1

1 of 3

$tt{(a) to solve this task we will use 2 formulas:\ for unpolarized light $I_f=frac{1}{2}I_i$}$
Law of Malus :$I_f=I_i cos^2 theta$
we can the deduce the following:

$$
I_f=I_2 cos^2 theta_2=I_1 cos^2 theta_2*cos^2 theta_1= frac{1}{2} I_i cos^2 theta_2 cos^2 theta_1=
frac{1}{2} (25.5frac{W}{m^2}) cos^2 45*cos^2 45=3.2 frac{W}{m^2}
$$

Step 2

2 of 3

(b) for $theta_1=0$

$$
I_f=frac{1}{2} I_i cos^2 theta_2 cos^2 theta_1=
frac{1}{2} (25.5frac{W}{m^2}) cos^2 45*cos^2 0=6.4 frac{W}{m^2}
$$

Result

3 of 3

$$
tt{(a)$3.2 frac{W}{m^2}$,(b)$6.4 frac{W}{m^2}$}
$$

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