Question

27. An analyser is inclined to a polariser at an angle
of 30°. The intensity of light emerging from the analyser is (1/n) thth of that is incident on the polariser. Then n is equal to
1) 4
2) 4/3
3)8/3
4) 1/4​

Answer 1

Solution :

Let $\sf{I_{0}}$ be the intensity of unpolarised light. Intensity after passing through the polariser,

$\implies$ $\huge{\boxed{\sf{I =\frac{I_{0}}{2}}}}$

Intensity of the light after passing through the analyser,

$\implies$ $\sf{I'=I\:cos^{2}30}$

That is:

$\implies$ $\sf{I'=\frac{I_{0}}{2}\:cos^{2}30}$

$\implies$ $\sf{I'= \frac{3I_{0}}{8}}$

And,

$\implies$ $\sf{I' = \frac{1}{n}\:I_{0}}$

$\implies$ $\sf{n = \frac{8}{3}}$

Therefore,

Correct option: (3) 8/3

__________________

Answered by: Niki Swar, Goa❤️

Answer 2

27. An analyser is inclined to a polariser at an angle

of 30°. The intensity of light emerging from the analyser is (1/n) thth of that is incident on the polariser. Then n is equal to

1) 4

2) 4/3

3)8/3

4) 1/4