Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University
9th Edition
ISBN: 9781305372337
Author: Raymond A. Serway | John W. Jewett
Publisher: Cengage Learning
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Chapter 38, Problem 48P

(a)

To determine

The fraction by which the transmitted intensity is reduced.

(a)

Expert Solution
Check Mark

Answer to Problem 48P

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)

To determine

The fraction by which intensity is reduced when 4 polarizing filters are placed.

(b)

Expert Solution
Check Mark

Answer to Problem 48P

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)

To determine

The fraction by which intensity is reduced when 7 polarizing filters are placed.

(c)

Expert Solution
Check Mark

Answer to Problem 48P

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)

To determine

The comparison between answer of part (a), (b) and (c).

(d)

Expert Solution
Check Mark

Answer to Problem 48P

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 38 Solutions

Physics For Scientists And Engineers With Modern Physics, 9th Edition, The Ohio State University

Ch. 38 - Prob. 4OQCh. 38 - Prob. 5OQCh. 38 - Prob. 6OQCh. 38 - Prob. 7OQCh. 38 - Prob. 8OQCh. 38 - Prob. 9OQCh. 38 - Prob. 10OQCh. 38 - Prob. 11OQCh. 38 - Prob. 12OQCh. 38 - Prob. 1CQCh. 38 - Prob. 2CQCh. 38 - Prob. 3CQCh. 38 - Prob. 4CQCh. 38 - Prob. 5CQCh. 38 - Prob. 6CQCh. 38 - Prob. 7CQCh. 38 - Prob. 8CQCh. 38 - Prob. 9CQCh. 38 - Prob. 10CQCh. 38 - Prob. 11CQCh. 38 - Prob. 12CQCh. 38 - Prob. 1PCh. 38 - Prob. 2PCh. 38 - Prob. 3PCh. 38 - Prob. 4PCh. 38 - Prob. 5PCh. 38 - Prob. 6PCh. 38 - Prob. 7PCh. 38 - Prob. 8PCh. 38 - Prob. 9PCh. 38 - Prob. 10PCh. 38 - Prob. 11PCh. 38 - Coherent light of wavelength 501.5 nm is sent...Ch. 38 - Prob. 13PCh. 38 - Prob. 14PCh. 38 - Prob. 15PCh. 38 - Prob. 16PCh. 38 - Prob. 17PCh. 38 - Prob. 18PCh. 38 - What is the approximate size of the smallest...Ch. 38 - Prob. 20PCh. 38 - Prob. 21PCh. 38 - Prob. 22PCh. 38 - Prob. 23PCh. 38 - Prob. 24PCh. 38 - Prob. 25PCh. 38 - Prob. 26PCh. 38 - Consider an array of parallel wires with uniform...Ch. 38 - Prob. 28PCh. 38 - Prob. 29PCh. 38 - A grating with 250 grooves/mm is used with an...Ch. 38 - Prob. 31PCh. 38 - Prob. 32PCh. 38 - Light from an argon laser strikes a diffraction...Ch. 38 - Show that whenever white light is passed through a...Ch. 38 - Prob. 35PCh. 38 - Prob. 36PCh. 38 - Prob. 37PCh. 38 - Prob. 38PCh. 38 - Prob. 39PCh. 38 - Prob. 40PCh. 38 - Prob. 41PCh. 38 - Prob. 42PCh. 38 - Prob. 43PCh. 38 - Prob. 44PCh. 38 - Prob. 45PCh. 38 - Prob. 46PCh. 38 - Prob. 47PCh. 38 - Prob. 48PCh. 38 - Prob. 49PCh. 38 - Prob. 50PCh. 38 - Prob. 51PCh. 38 - Prob. 52PCh. 38 - Prob. 53APCh. 38 - Prob. 54APCh. 38 - Prob. 55APCh. 38 - Prob. 56APCh. 38 - Prob. 57APCh. 38 - Prob. 58APCh. 38 - Prob. 59APCh. 38 - Prob. 60APCh. 38 - Prob. 61APCh. 38 - Prob. 62APCh. 38 - Prob. 63APCh. 38 - Prob. 64APCh. 38 - Prob. 65APCh. 38 - Prob. 66APCh. 38 - Prob. 67APCh. 38 - Prob. 68APCh. 38 - Prob. 69APCh. 38 - Prob. 70APCh. 38 - Prob. 71APCh. 38 - Prob. 72APCh. 38 - Prob. 73APCh. 38 - Light of wavelength 632.8 nm illuminates a single...Ch. 38 - Prob. 75CPCh. 38 - Prob. 76CPCh. 38 - Prob. 77CPCh. 38 - Prob. 78CPCh. 38 - Prob. 79CP
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