Question

A.

A laser beam has a power of 10 W of blue light that is horizontally polarized. It passes through a series of filters with axis of polarization at 0° w.r.t. the horizontal, 45° w.r.t. the horizontal, and 90° w.r.t. the horizontal. What is the power that emerges from these polarizers? Optic axis 2 = 45° Optic axis =90° Eof incoming light Optic axis 1=0° O 2.5 W Oow 01.25 W O 5W

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C.

D.

Answer 1

A) When an unpolarized light incidents on a polarizer, the power of polarized light coming out of the polarizer is half of the power of unpolarized incident light, that is; I1 = I0/2
Here, I0 = 10 W
After passing through the first filter with axis of polarization at 0o w.r.t. the horizontal, the power of polarized light, I1 = 10 / 2
= 5 W

Now the polarized light having 5W power will pass through the second filter with axis of polarization at 45o w.r.t. the horizontal.
The power of the polarized light becomes I2 = I1 cos2$\theta$
= 5 x cos2(45)
= 2.5 W

Now this polarized light will pass through the third filter with axis of polarization at 90o w.r.t. the horizontal. Compared to the previous polarizor, the optical is now 45o.
The power of the of the polarized light emerging from these three filters, I3 = I2 cos2$\theta$
= 2.5 x cos2(45)
= 1.25 W