Please answer parts A B C D and E. Class Management I Help HW v2 ch 16 & v3 ch 1

Question

Please answer parts A B C D and E.

Class Management I Help HW v2 ch 16 & v3 ch 1.7 Begin Date: 4/26/2017 12:00:00 AM Due Date: 5/4/2017 11:59:00 PM End Date 5/4/2017 11:59:00 PM (13%) Problem 7: Unpolarized light of intensity To 1300 Wim is incident upon three polarizers. The axis of the first polarizer is vertical. The axis of the second polarizer is rotated at an angle 6 75 from the vertical. The axis of the third polarizer is horizontal Randomized Variables 1300 W/m2 Otheexpertta.com A 20% Part (a) What is the intensity of the light in Wim after it passes through the first polarizer Grade Summary Deductions 100% Potential sino coso tano 8 9 Submission cotano as 4 5 6 Attempts remaining: acos0 (0% per attempt 0 1 2 3 atan0 acotan0 sinh detailed view cosh0 tanh0 cotanh0 Degree O Radians Submit Hint Hints 0% deduction per hint. Hints remaining Feedback: 0% deduction per feedback. A 20% Part (b) What is the intensity of the light in W/m3 after it passes through the second polarizer? What is the intensity in W m of the light after it passes through the third polarizer 4 20% Part (d Write an equation for the intensity of light after it has passed through all three polarizers in terms of the intensity of unpolarized light entering the first polarizer Io and the angle of the second polarizer relative to the first, given that the first and third polarizers are crossed (009 between them). Use trigonometric identities to simplify and give the results in terms of a single trigonometric function of o 26 A 20% Part (e) What is the smallest angle 62, in degrees, from the vertical that the second polarizer could be rotated such that the intensity of light passing through all three polarizer is 13 12 W/m s

Explanation / Answer

here,

I0 = 1300 W/m^2

A)

the intensity of light passing throught the first polarizer , I1 = I0 /2 ....(1)

I1 = 1300 /2 = 650 W/m^2

B)

the intensity of light when it passes through the seccond polarizer , I2= I1 * ( cos(75))^2 ....(2)

I2 = 650 * cos^2(75)

I2 = 43.5 W/m^2

C)

the intensity of light when it passes through the third polarizer , I3 = I2 * ( cos(90 - 75))^2 ....(3)

I3 = 43.5 * cos^2(15)

I3 = 40.59 W/m^2

D)

from (1) , (2) and (3)

I3 = 0.5 * I0 * ( cos(theta))^2 * ( cos(90 - theta))^2

I3 = 0.5 * I0 * ( cos(theta) * sin(theta))^2

I3 = 0.5 * I0 * ( sin(2*theta) /2)^2

I3 = 0.125 * I0 * sin(2*theta)

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