The answer is given in the book, However I am notunderstanding how they come up

Question

The answer is given in the book, However I am notunderstanding how they come up with this solution. I thought it was I=I" cos^2 (angle). Please help me tounderstand how they get 14 W/m Suppose that unpolarized light ofintensity 150 W/m2falls on the polarizer in the figure below, and the angle in the drawing is 30°. What is the light intensity reaching thephotocell? The answer is given in the book, However I am notunderstanding how they come up with this solution. I thought it was I=I" cos^2 (angle). Please help me tounderstand how they get 14 W/m Suppose that unpolarized light ofintensity 150 W/m2falls on the polarizer in the figure below, and the angle in the drawing is 30°. What is the light intensity reaching thephotocell?

Explanation / Answer

Given :
.           Intensity of polarised light I = 150  W /m2 .                      = 30o . Intensity after passing through polariser is: .                    I1 = I / 2 = 150 / 2 = 75 W/ m2 . Intensity after passing through insert is : .                   I2 = I1 cos 2            .                       = 75 * cos2 30   = 56.25 W /m2 . Intensity after passing through analyser I3 =I2 cos 2 ( 90 - ) .                                                               = 56.25 * cos2 (60)  = 14.06 W / m2 . Hope this helps u!                                                                              

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