A beam of partially polarized light (a mixture of horizontaly polarized and unpo

Question

A beam of partially polarized light (a mixture of horizontaly polarized and unpolarized light) is incident on a polarizing sheet. In order to determine the relative intensities of the polarized and unpolarized portions of the beam, the light is sent through a vertical polarizer, and the intensity is noted to drop by 80%.

a)Determine the intensities of the polarized and unpolarized portions of the beam.

b)what is the transmitted intensity if the polarizer is oriented horizontally?

c)at what angle will the transmitted intesity be the maximum?

d)if the intial beam was a mixture of vertically polarized and unpolarized light and the intensity dropped to 60% after passing through a horizontal polarizer, what would be the intensities of the different portions of the beam?

PLEASE HELP...DETAILS

Explanation / Answer

Let, Total intensity be I.

fraction of polarised light be x, and unpolarised be (1-x).

On passing through vertical polarizer,

Net intensity = xI * cos^90 + (1-x)I*0.5 =0.5I - 0.5xI

Given, 0.5I - 0.5xI = 0.2I

Thus, 0.5x = 0.3

x = 0.6

Thus, original intensity of polarized beam = 0.6 I

original intensity of unpolarized beam = 0.4 I

b) after passing through horizontal polarizer, angle = 0

Net intensity = 0.6*I*cos^0 + 0.4I / 2 = 0.8 I i.e. 80% of original intensity

c) To maximise transmitted intensity, angle = 0 degrees

d) xI * cos^0 + (1-x)I / 2 = 0.6 I

    xI + 0.5I - 0.5xI = 0.6I

    0.5 + 0.5x = 0.6

    x = 0.2

    Thus, polarized component = 20%

    unporized = 80%

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