Intensity of polaroids at an angle

[Sasha26]

Homework Statement

At what angle should the axes of two Polaroids be placed so as to reduce the intensity of the incident unpolarized light by an additional factor (after the first Polaroid cuts it in half) of 7? b) 25? c) 250?

Homework Equations

The first polaroid decreases the intensity by 1/2, so that's I1 = 0.5I0.
I also have I2 = I1(cos theta)^2.

The Attempt at a Solution

I think it might just be the wording that's getting me, but here's my interpretation:

In the first polaroid, 1/2 the intensity is removed. That makes I1 equal to 0.5 I0.
In the second polaroid, 1/7 of I0 is removed. That makes I2 = 0.5I0 - (1/7 I0), or 5/14 I0.

So then I've got I2 = I1(cos theta)^2 --> 5/14 I0 = 0.5 I0 cos(theta)^2. This gives me 32 as theta, which is incorrect. I haven't tried to work out the rest as there seems to be something wrong with my procedure and they're all essentially the same problem.

Any help is much appreciated!

 

[Redbelly98]

To reduce I1 by a factor of 7 means that

I2 = (1/7) I1

 

[thomate1]

Your interpretation is correct, but there is an error in your calculation for I2. It should be I2 = I1/7 = (0.5 I0)/7 = 0.0714 I0. Then, using the equation I2 = I1(cos theta)^2, we get 0.0714 I0 = 0.5 I0 cos(theta)^2. Solving for theta, we get cos(theta) = sqrt(0.0714/0.5) = 0.2673. Taking the inverse cosine, we get theta = 75.2 degrees. This is the angle at which the axes of the two polaroids should be placed to reduce the intensity of unpolarized light by a factor of 7 after the first polaroid cuts it in half.

For b) 25 degrees, we get cos(theta) = sqrt(0.0714/0.5) = 0.5345. Taking the inverse cosine, we get theta = 57.9 degrees. This is the angle at which the axes of the two polaroids should be placed to reduce the intensity by a factor of 7.

For c) 250 degrees, we get cos(theta) = sqrt(0.0714/0.5) = -0.5345. Taking the inverse cosine, we get theta = 122.1 degrees. This is also the angle at which the axes of the two polaroids should be placed to reduce the intensity by a factor of 7.