Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degrees, respectiv

Question

Three ideal polarizing filters are stacked, with the polarizing axis of the second and third filters at 21 degrees and 61 degrees, respectively, to that of the first. If unpolarized light is incident on the stack, the light has intensity 60.0 w/cm^ 2 after it passes through the stack.
If the incident intensity is kept constant:
1) What is the intensity of the light after it has passed through the stack if the second polarizer is removed?
2) What is the intensity of the light after it has passed through the stack if the third polarizer is removed?

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Philomena 4 years 2021-07-15T07:35:36+00:00 1 Answers 0 views 0

Answers ( )

  1. Ladonna
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    2021-07-15T07:36:56+00:00
    Reply

    Answer:

    1

    When second polarizer is removed the intensity after it passes through the stack is    

                        I_f_3 = 27.57 W/cm^2

    2 When third  polarizer is removed the intensity after it passes through the stack is    

                    I_f_2 = 102.24 W/cm^2

    Explanation:

      From the question we are told that

           The angle of the second polarizing to the first is  \theta_2 = 21^o  

            The angle of the third  polarizing to the first is     \theta_3 = 61^o

            The unpolarized light after it pass through the polarizing stack   I_u = 60 W/cm^2

    Let the initial intensity of the beam of light before polarization be I_p

    Generally when the unpolarized light passes through the first polarizing filter the intensity of light that emerges is mathematically evaluated as

                         I_1 = \frac{I_p}{2}

    Now according to Malus’ law the  intensity of light that would emerge from the second polarizing filter is mathematically represented as

                        I_2 = I_1 cos^2 \theta_1

                           = \frac{I_p}{2} cos ^2 \theta_1

    The intensity of light that will emerge from the third filter is mathematically represented as

                      I_3 = I_2 cos^2(\theta_2 - \theta_1 )

                              I_3= \frac{I_p}{2}(cos^2 \theta_1)[cos^2(\theta_2 - \theta_1)]

    making I_p the subject of the formula

                      I_p = \frac{2L_3}{(cos^2 \theta [cos^2 (\theta_2 - \theta_1)])}

        Note that I_u = I_3 as I_3 is the last emerging intensity of light after it has pass through the polarizing stack

             Substituting values

                          I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (61-21)])}

                          I_p = \frac{2 * 60 }{(cos^2(21) [cos^2 (40)])}

                               =234.622W/cm^2

    When the second    is removed the third polarizer becomes the second and final polarizer so the intensity of light would be mathematically evaluated as

                          I_f_3 = \frac{I_p}{2} cos ^2 \theta_2

    I_f_3 is the intensity of the light emerging from the stack

                         

    substituting values

                         I_f_3 = \frac{234.622}{2} * cos^2(61)

                           I_f_3 = 27.57 W/cm^2

      When the third polarizer is removed  the  second polarizer becomes the

    the final polarizer and the intensity of light emerging from the stack would be  

                      I_f_2 = \frac{I_p}{2} cos ^2 \theta_1

    I_f_2 is the intensity of the light emerging from the stack

    Substituting values

                      I_f_2 = \frac{234.622}{2} cos^2 (21)

                         I_f_2 = 102.24 W/cm^2

       

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