Question

Two polarizing sheets have their transmission axes cross so no light can pass through. I place a third sheet in with it axes at an angle ? with respect to the first sheet. (a) Derive an expression for the intensity of transmitted light through all three sheets as a function of ?. (b) Find the angle for which the transmitted intensity is maximized?

Answer 1

Unpolarized light of intensityI₀ is incident on the first sheet.
so through first sheet, intensity is given by malus rule.

I₁ =0.5I₀
through first two sheets, intensity is hiven by.

I₂= 0.5I₀ *cos² _theta
through all three sheets, intensityI₃ = 0.5I₀ *cos²theta*cos²(90 - theta)
a) theta = 45^o
I₁ = 0.5I₀
I₂ = 0.5I₀ *cos²(45) = 0.25I₀
I₃ = 0.5I₀ *cos²(45) * cos²(45) = 0.125I₀

B.

I3 = I1 * cos^2(theta)*sin^2(theta)

If theta = 45°

Then I3/I1 = 0.25.

Hence at theta = 45 the value will be maximum.