First slide

Diffraction of light-Wave Optics

Question

An analyser is inclined to a polariser at an angle of $30^{0}$ . The intensity of light emerging from the analyser is $\frac{1}{n}^{^{t h}}$ of that is incident on the polarizer. Then n is equal to

Easy

A

$4$
B

$\frac{4}{3}$
C

$\frac{8}{3}$
D

$\frac{1}{4}$

Solution

Here , $θ = 30^{0}$ , $I = \frac{I_{0}}{n}$
Intensity of polarized light $I = \frac{I_{0}}{2} \cos^{2} θ$

                                           $\frac{I_{0}}{n} = \frac{I_{0}}{2} \cos^{2} 30$

                                                 $\Rightarrow \frac{I_{0}}{2} . \frac{3}{4}$

                                         $\frac{I_{0}}{n} = \frac{3 I_{0}}{8}$

                                       $∴ n = \frac{8}{3}$

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