1.A polarizer is placed in front of a source of unpolarized light. The intensity

Question

1.A polarizer is placed in front of a source of unpolarized light. The intensity after passing through the polarizer is 1 mW. What was the intensity before passing through the polarizer?

a) 4 mW b) 2 mW c) 1mW d) 1/2 mW e) 1/4 mW

2.The beam from question 1 then passes through a second polarizer which is rotated 40° relative to the first polarizer. What is the intensity of the beam after passing through both polarizers?
a) 0.766 mW b) 0.643 mW c) 0.587 mW d) 0.413 mW e) none of the above

3.A beam of light originally consists of two wavelengths 417 nm and 590 nm in equal intensity. After undergoing Rayleigh scattering how much brighter is the 417nm wavelength?
a) 2 times b) 4 times c) 8 times d) 16 times e) 32 times

4.Light reflected off a horizontal surface will be
a) partially polarized with the plane of polarization vertical b) partially polarized with the plane of polarization horizontal c) completely polarized
d) completely unpolarized

5.What is the Brewster's angle for light traveling from air to crown glass?

a) 48.9° b) 41.1° c) 33.3° d) 56.7° e) none of the above

Explanation / Answer

Part 1

when unpolarised light pass through polariser intensity of light become half of initial intensity. here after polarisation intensity is

I = 1mW

so befire polariosation intensity is

Io = 2*I = 2mW

option B

Part 2

Intensity of light after passing through second polariser

I = Io cos^2(theta)

I = 1 * cos^2(40)

I = 0.587 mW

option C

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