Question

Unpolarized light passes through a succession of three polarizers. Their orientations with respect to the vertical are 0?, 60 ?, and 45 ?, respectively.

Part A: Find the fraction of the original light intensity remaining after first polarizer.

Part B: Find the fraction of the original light intensity remaining after second polarizer.

Part C: Find the fraction of the original light intensity remaining after third polarizer.

Answer 1

When light passes through first polarizer, it's intensity will be reduced to half, since the light is unpolarized

I1 = I0/2

I1/I0 = 1/2 = 0.5

After that when light passes thorugh 2nd polarizer, Intensity will be from Malus's law

I2 = I1*(cos A1)^2

A1 = 60 deg

I2 = (I0/2)*(cos 60 deg)^2

I2/I0 = 1/8 = 0.125

After that when light passes through 3rd polarizer, Intensity will be

I3 = I2*(cos A2)^2

A2 = 45 - 60 = -15 deg

I3 = (I0/8)*(cos -15 deg)^2

I3/I0 = (1/8)*(cos -15 deg)^2 = 0.117

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