Question

A beam of unpolarised light with intensity I0 falls first upon a polarizer with transmission axis ?TA,1, then upon a second polarizer with transmission axis ?TA,2, where ?TA,2 – ?TA,1 = 90 degrees (in other words the two axes are perpendicular to one another).

a. What is the intensity I2 of the light beam emerging from the second polarizer? Show your working.

Now a third polarizer (Polarizer A in the figure below), whose transmission axis is offset 45 degrees from each of the others, is inserted between the original two polarizers.

b. What is the new intensity I2 of the light emerging from the final polarizer?

Answer 1

a)

using malus law

the intensity I can be taken as

\(\mathrm{I}=\mathrm{I}_{0} \cos ^{2}(\theta)\)

OTA,

\(2-\theta T A\)

\(1=90\)

here given

\(\theta=90^{\circ}\)

\(I_{2}=I_{0} X \cos (90)\)

\(I_{2}=0\)

the fianl intensity is \(I_{2}=0\)

b)

for the third polarizer given is

\(\theta=45^{\circ}\)

again considering the malus law

\(I_{2}=I_{0} \cos ^{2} 45\)

\(I_{2}=I_{0} \times 0.5\)

\(I_{2}=I_{0} / 2\)