00:01
In the first part of this problem, we have to find the value of phi for which the given condition holds.
00:09
So the intensity of light when it passes through the first polarizer is written as ipriam equals to 1 over 2 of i0, where this i note is the intensity of unpolarized light.
00:22
When the light passes through the second polarizer, its intensity will be i equals to i prime, cosine, square of, of phi.
00:33
Now inserting the value of i and i prime we can write this equation as i0 divided by 10 equals to i not divided by 2 cosine square of phi.
00:48
Well now solving this equation for phi we can write it is phi equals to arc cosine of 1 divided by square root of 5.
01:00
So this will give you at the value for phi as phi equals to to 63 .43 degree.
01:14
In part b of this problem, we have to calculate the value of the phi, which holds the given condition.
01:23
So there will be no change in the intensity of light after passing through the first polarizer...