PHYSICS:F/SCI.+ENGRS.,V.1
PHYSICS:F/SCI.+ENGRS.,V.1
10th Edition
ISBN: 9781337553575
Author: SERWAY
Publisher: CENGAGE L
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Chapter 37, Problem 32P

An unpolarized beam of light is incident on a stack of ideal polarizing filters. The axis of the first filter is perpendicular to the axis of the last filter in the stack. Find the fraction by which the transmitted beam’s intensity is reduced in the three following cases. (a) Three filters are in the stack, each with its transmission axis at 45.0° relative to the preceding filter. (b) Four filters are in the stack, each with its transmission axis at 30.0° relative to the preceding filter. (c) Seven filters are in the stack, each with its transmission axis at 15.0° relative to the preceding filter. (d) Comment on comparing the answers to parts (a), (b), and (c).

(a)

Expert Solution
Check Mark
To determine
The fraction by which the transmitted intensity is reduced.

Answer to Problem 32P

The intensity will be reduced by 0.875 fraction of the incident intensity.

Explanation of Solution

Given info: The number of the polarizing filters is 3 . The formula to calculate the intensity after it passes through any polarizing filter is,

I=I0cos2(θ) . (1)

Here,

I is the intensity that came out of the polarizing filter.

I0 is the incident intensity of the un polarized light.

θ is the angle between the transmission of the two polarizing filters.

When an unpolarized light is passed through a polarizing filter intensity is reduced to half. So after passing through the first polarizer the intensity of the light becomes half.

I01=I02

Here,

I01 is the intensity of the light after the first polarizing filter

The angle between the transmission axis of second polarizer and the first polarizer is 45.0° . Therefore, from equation (1) the formula to calculate the intensity when the light comes out of the second polarizer is,

I02=I01(cosθ)2 (2)

Here,

I02 is the intensity of the light after the second  polarizing filter

The third polarizing filter and the second polarizing filter has the same 45.0° angle in between their transmission axis is also

Therefore the final intensity after three polarizing filters is,

I03=I02(cosθ)2 (3)

Substitute I01(cosθ)2 for I02 in equation (3),

I03=I02(cosθ)2I03=I01(cosθ)2(cosθ)2 (4)

Substitute I02 for I01 in equation (4),

I03=I01(cosθ)2(cosθ)2I03=I02(cosθ)2(cosθ)2 (5)

From equation (5), a general formula for the calculation of intensity when light is passed through n number of polarizing filters is,

I=I02(cos2θ)n1 (6)

Here,

I is the final intensity.

I0 is the initial intensity.

n is the number of polarizing filters.

Substitute 45.0° for θ in equation 5 to calculate the fraction of intensity transmitted after three polarizing filters,

I03=I02(cosθ)2(cosθ)2=I02(cos45°)2(cos45°)=I02(12)(12)=0.125I0

Therefore the absorbed intensity is

Iabs=I00.125Io=0.875I0 (7)

Conclusion:

Therefore, the fraction by which the intensity is reduced is 0.875 of the incident intensity.

(b)

Expert Solution
Check Mark
To determine
The fraction by which intensity is reduced when 4 polarizing filters are placed.

Answer to Problem 32P

The fraction by which the intensity is reduced is 0.789 of the incident intensity.

Explanation of Solution

Given info: The number of filters are 4 and the angle between the transmission axis is 30.0° .

From equation (6) the formula to calculate when there are n number of polarizing filters are present,

I=I02(cos2θ)n1

Substitute 30.0° for θ and 4 for n in the above equation,

I=I02(cos2θ)n1I=I02(cos230.0°)41I=I0(0.211)

Therefore the absorbed intensity is

Iabs=I00.211I0=0.789I0 (8)

Conclusion:

Therefore, The fraction by which the intensity is reduced is 0.789 of the incident intensity

(c)

Expert Solution
Check Mark
To determine
The fraction by which intensity is reduced when 7 polarizing filters are placed.

Answer to Problem 32P

The fraction by which the intensity is reduced is 0.670 of the incident intensity.

Explanation of Solution

Given info: The number of filters are 7 and the angle between the transmission axis is 15.0° .

From equation (6) the formula to calculate when there are n number of polarizing filters are present,

I=I02(cos2θ)n1

Substitute 15.0° for θ and 7 for n in the above equation,

I=I02(cos2θ)n1I=I02(cos215.0°)71I=I0(0.330)

Therefore the absorbed intensity is

Iabs=I00.330I0=0.670I0 (9)

Conclusion:

Therefore, the fraction by which the intensity is reduced is 0.670 of the incident intensity

(d)

Expert Solution
Check Mark
To determine
The comparison between answer of part (a), (b) and (c).

Answer to Problem 32P

The intensity of light can be increased by increasing the number of stacks of polarizing filters by decreasing the angle between their transmission axis.

Explanation of Solution

From equation (7), (8) and (9), it is evident that, as the number of polarizing filters increased the fraction of absorbed was decreased. For the case of 3 polarizing filters and angle between the transmission axis 45.0° the absorbed intensity was 0.875 of the incident intensity, for the case of 4 filters and angle between the transmission axis 30.0° the absorption was reduced to 0.789 of the incident intensity and finally for the case of 7 filters and angle between the transmission axis 15.0° the absorption was only 0.670 of the incident intensity.

Conclusion:

Therefore, the intensity of light can be increased by increasing the number of stacks of polarizing filters by decreasing the angle between their transmission axis.

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Chapter 37 Solutions

PHYSICS:F/SCI.+ENGRS.,V.1

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