Physics: 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 is C. Kelvin, Kelvin is the temperature scales does water boil at the highest numerical value. Hope it helped you, and have a great day.-Charlie...

Unpolarized light with an intensity of 655 w / - id65524322726, micahatwood03

Ответ:

BaileyElizabethRay

13.01.2022

Physics

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

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

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

$\theta=60.15^{0}$

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