The intentsity out of two polarizers

[Rahulrj]

Homework Statement

A vertically polarized light of intensity $I_0$ is incident on a polarizer whose axis makes an angle $\theta$ to the vertical. The light then passes through another polarizer whose axis make 60 seg to that of the first one. The net intensity of the output light in relation to incident intensity is
a) $I_0 \cos^2\theta$
b) $2I_0 \cos^2\theta$
c) $3/2I_0 \cos^2\theta$
d) 0

Homework Equations

$I =I_0 \cos^2\theta$

The Attempt at a Solution

This was supposed to be an easy one but I don't get any of the answers given in the options and I am totally confused about where I made the mistake. so this is how I did:
through the first polarizer $I_1 = I_0 \cos^2\theta$
through the second one $I_2 = I_0 \cos^2\theta \cos^2(60)$ the question says the axis of the second one is at 60 deg to that of first one hence the light coming from the first must also make an angle at 60 right?

 

[kuruman]

You are correct. The output intensity should be as you say with $\cos^2(60^o) = \frac{1}{4}$.

 

[Rahulrj]

Then the questioner probably meant that the axis of the second polarizer is at 60 degrees to the vertical which would give an angle of 30 deg between the light and the axis of the second polarizer which in turn will give the answer $3/2 I_0 \cos^2\theta$. option c.

 

[kuruman]

It is risky to try and read the minds of questioners. Besides, $\cos^2(30^o) = \frac{3}{4}$, not $\frac{3}{2}$.

 

[Rahulrj]

It is risky to try and read the minds of questioners. Besides, $\cos^2(30^o) = \frac{3}{4}$, not $\frac{3}{2}$.

Oh yes it is 3/4 but the answer is still close to option c. It made me much frustrated to see I didn't get anyone of the answers and consumed lot of my time going through all the relevant concepts and examples to find out where I have gone wrong! Thank you anyway!

 

[kuruman]

Oh yes it is 3/4 but the answer is still close to option c.

Maybe close but still incorrect. Look at it this way: after the first polarizer, the intensity is I₁ = I₀cos²θ. After the second polarizer, I₂ should be less than I₁, but not zero because the second polarizer is not at 90^o w.r.t. to the first. None of the 4 answers fit these constraints.