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Plane-polarized light is incident on a single polarizing disk with the direction of E0 parallel to the direction of the transmission axis. Through what angle should the disk be rotated so that the intensity in the transmitted beam is reduced by a factor of:

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Nandhitha1098

Answer: Plane-polarized light is incident on a single polarizing disk with the direction of E0 parallel to the direction of the transmission axis. Through angle [tex]\alpha=cos^-1(\sqrt{\frac{1}{n} })[/tex] should the disk be rotated so that the intensity in the transmitted beam is reduced by a factor of n .

Explanation: To find the correct answer, we need more clarifications about the Malu's law.

What is Malu's law?

  • When a beam of plane polarized light is passed through an analyzer, the intensity I of the transmitted light varies directly as the square of the angle between the direction of polarizer and the analyzer.

                                 [tex]I=I_0cos^2\alpha[/tex]

where, [tex]I_0[/tex] is the maximum intensity of transmitted light.

  • Intensity of the light transmitted through the polarizer is half of the incident intensity.

How to solve the problem?

  • Here in the question ,let's take the factor at which the intensity of transmitted light is reduced by as n.
  • we have the expression

                                            [tex]I=I_0cos^2\alpha[/tex]

  • we have to find the value of angle alpha, so that the intensity in the transmitted beam is reduced by a factor of n.

                                 [tex]\alpha =cos^-1\sqrt{(\frac{I}{I_0}} )=cos^-1\sqrt{\frac{I}{nI} } \\\alpha =cos^-1\sqrt{\frac{1}{n} }[/tex]

  • If we know the value of n, then we can easily find the angle.

Thus, we can conclude that, the value of angle alpha, so that the intensity in the transmitted beam is reduced by a factor of n will be,

                                 [tex]\alpha=cos^-1(\sqrt{\frac{1}{n} })[/tex] .

Learn more about Malu's law here: https://brainly.com/question/28045350

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