PLEASE ANSWER ALL PARTS, THANK YOU! PLEASE ANSWER ALL PARTS, THANK YOU! (3%) Pro

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PLEASE ANSWER ALL PARTS, THANK YOU!

PLEASE ANSWER ALL PARTS, THANK YOU! (3%) Problem 19: Linearly polarized light is transmitted through two polarizers as shown in the figure. The light is polarized in they direction. The axis of the first polarizer is at theta1 = 28 to the y direction- The axis of the second polarizer is at theta2 = 54 to they direction. The intensity of the initial beam of polarized light is Imax,. Randomized Variables Polarized Light Theat1 = 28o Theta2 = 54o & 25% Part (a) Express the intensity of the light after it passes through the first polarizer, I1, in terms of I0, and theta1. & 25% Part (b) Express the intensity of the light after it passes through the second polarizer, I2, in terms of I1 delta theat12, where delta theta12 = theta2 - theta1. & 25% Part (c) 1ff is the fraction of the incident light transmitted after it passes through the two polarizers, express f in terms of theat1 and delta theta12, where delta theta12 = theta2 - theta1. & 25% Part (d) Solve for the numerical value of f. (3%) Problem 19: Linearly polarized light is transmitted through two polarizers as shown in the figure. The light is polarized in they direction. The axis of the first polarizer is at theta1 = 28 to the y direction- The axis of the second polarizer is at theta2 = 54 to they direction. The intensity of the initial beam of polarized light is Imax,. Randomized Variables Polarized Light Theat1 = 28 degree Theta2 = 54 degree & 25% Part (a) Express the intensity of the light after it passes through the first polarizer, I1, in terms of I0, and theta1. & 25% Part (b) Express the intensity of the light after it passes through the second polarizer, I2, in terms of I1 delta theat12, where delta theta12 = theta2 - theta1. & 25% Part (c) 1ff is the fraction of the incident light transmitted after it passes through the two polarizers, express f in terms of theat1 and delta theta12, where delta theta12 = theta2 - theta1. & 25% Part (d) Solve for the numerical value of f.

Explanation / Answer

Here ,

theta1 = 28 degree

theta2 = 54 degree

part a)

the output intensity after the first polarizer

I1 = Io * cos^2(theta1)

part b)

I2 = I1 * cos^2(theta12)

part c)

I2 = I1 * cos^2(theta12)

I2 = Io * cos^2(theta1) * cos^2(theta12)

f = cos^2(theta1) * cos^2(theta12)

part d)

f = cos^2(28) * cos^2(54 - 28)

f = 0.629

the fraction f is 0.629

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