Question

A series of polarizers are each placed at a 12 ∘ interval from the previous polarizer. Unpolarized light is incident on this series of polarizers. Part A How many polarizers does the light have to go through before it is below 110 of its original intensity?

Answer 1

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

I1 = I0/2

After that when light passes through 2nd polarizer, Intensity will be

I2 = I1*(cos A1)^2 = (I0/2)*(cos A1)^2

Now Suppose n polarizers are required for final intensity to be 1/10 th of original intensity, then

In = I0/10

Also intensity of light after passing through n polarizer will be:

In = (I0/2)*(cos A1)^(2*(n - 1))

here power of cosine is (n - 1) and not 'n' because after passing through 1st polarizer intensity will be dropped to half, after that it will reduce with (cos A1)^2, so remaining polarizer will be (n - 1)

Since In = I0/10, And each polarizer is placed at 12 deg interval, So

A1 = 12 deg

I0/10 = (I0/2)*(cos 12 deg)^(2*(n - 1))

1/5 = (cos 12 deg)^(2*(n - 1))

take logarithm on both sides

ln (1/5) = 2*(n - 1)*ln (cos 12 deg)

n = 1 - ln 5/(2*ln cos 12 deg)

n = 37.42 polarizers

So number of polarizers required will be 38 (next integer value)

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