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An analyser is inclined to a polariser a...

An analyser is inclined to a polariser at an angle of `30^(0)` . The intensity of light emerging from the analyser is `(1)/(n)` th of that is incident on the polariser. Then n is equal to

A

4

B

`4//3`

C

`8//3`

D

`1//4`

Text Solution

AI Generated Solution

To solve the problem, we will use Malus's Law, which states that the intensity of light passing through a polarizer is given by: \[ I = I_0 \cos^2(\theta) \] where: - \( I \) is the intensity of light after passing through the polarizer, - \( I_0 \) is the intensity of the incident light, - \( \theta \) is the angle between the light's polarization direction and the axis of the polarizer. ...
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