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

1.
`\theta=29.84^(0)`

2.
`\theta=60.15^(0)`

Step-by-step 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=(I_(0) )/(2)`

`I= (655(W)/(M^(2) ) )/(2)`

`I=327.5(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`

`(I_(2) )/(I_(1) )=Sin^(2)\theta`

taking square root on both sides

`\sqrt{(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`

`(I_(2) )/(I_(1) )=Cos^(2)\theta`

taking square root on both sides

`\sqrt{(163)/(327.5) } = cos\theta`

`\theta=Cos^(-1)(0.4977)`

`\theta=60.15^(0)`