Question

Unpolarized light with an intensity of 655 W / m2 is incident on a polarizer with an unknown axis. The light then passes through a second polarizer with has an axis which makes an angle of 81.5° with the vertical. After the light passes through the second polarizer, its intensity has dropped to 163 W / m2. What angle does the first polarizer make with the vertical?

Answer 1

Answer:

1.$\theta=29.84^{0}$

2.$\theta=60.15^{0}$

Explanation:

Polarizes axis can create two possible angles with the vertical.

first we have to find the intensity of  first polarizer

which is given as

$I=\frac{I_{0} }{2}$

$I= \frac{655\frac{W}{M^{2} } }{2}$

$I=327.5\frac{W}{m^{2} }$

For a smaller angle for the first polarizer:

According to Malus Law

$I_{2} =I_{1} Cos^{2}(90^{0} - \theta)$

$I_{2} =I_{1} sin^{2}\theta$

$\frac{I_{2} }{I_{1} }=Sin^{2}\theta$

taking square root on both sides

$\sqrt{\frac{163}{327.5} } = sin\theta$

$\theta=Sin^{-1}(0.4977)$

$\theta=29.84^{0}$

For a larger angle for the first polarizer:

According to Malus Law

$I_{2} =I_{1} cos^{2}\theta$

$\frac{I_{2} }{I_{1} }=Cos^{2}\theta$

taking square root on both sides

$\sqrt{\frac{163}{327.5} } = cos\theta$

$\theta=Cos^{-1}(0.4977)$

$\theta=60.15^{0}$