https//usq45tx.theexpertta.com/Common/TakeTutorialAssignment.aspx Class Manageme
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https//usq45tx.theexpertta.com/Common/TakeTutorialAssignment.aspx Class Management I Help Hw33- Electromagnetic Waves Begin Date: 5/3 0/2017 12:00:00 AM- Due Date: 8/1/2017 12:00:00 AM End Date: 8/15/2017 12:00:00 AM 1 AM End Date: 8/15/2017 12:00:00 AM (9%) Problem 13: Unpolarized light of intensity 10-950 wm2 is incident upon three polarizers. The axis of the first polarizer is vertical. The axis of the second polarizer is rotated at an angle 45from the vertical. The axis of the third polarizer is horizontal. Randomized Variables o-950 Wm2 -45° Ctheexpertta.com 20% Part (a) What is the intensity of the light in w m2 after it passes through the first polarizer? pleted pleted 1-20% pleted npleted mpleted upleted mpleted op 20% Part (b) what is the intensity of the light in Wan, after it passes through the second polarizer? ü-20% Part (c) What is the intensity in w mr of the light after it passes through the third polarizer? 20% Part (d) Wite an equation for the intensity of light after it has passed through all three polarzers in terms of the intensity of unpolarized light entering the first polarizer lo and the angle of the second polarizer relative to the first, given that the crossed (90° between them). Use trigonometrie identities to sinmplify and give the results in tenns of a single trigonometric fixstion of o 28 20-, Part (e) Wat is the smallest angle -m degrees from the vertical that the second polarizer could be rotated such that the atensity of light passing through all three polarizers is l' wm2? Grade Summary Potential Submissions 100% mpleted mpleted sino cos)Explanation / Answer
After polarizer 1, I=650W/m2
After polarizer 2, inclined at 45° to this, I=650cos245° =325 W/m2
After polarizer 3, I=66=325cos2theta
So, theta =63.21°
Initially, polarizer 2 and 3 are inclined at 45° to each other, so additional inclination required is 63.21-45= 18.21°
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