(D) The angle of polarisation $i_{p}$ is given by Brewster's law: $\mu = \tan i_{p}$.Given $i_{p} = 60^{\circ}$,the refractive index $\mu = \tan 60^{\

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$\sin ^{-1} \frac{1}{\sqrt{3}}$

(D) The angle of polarisation $i_{p}$ is given by Brewster's law: $\mu = \tan i_{p}$.Given $i_{p} = 60^{\circ}$,the refractive index $\mu = \tan 60^{\circ} = \sqrt{3}$.The critical angle $C$ is related to the refractive index by the formula: $\sin C = \frac{1}{\mu}$.Substituting the value of $\mu$,we get $\sin C = \frac{1}{\sqrt{3}}$.Therefore,the critical angle $C = \sin ^{-1} \left( \frac{1}{\sqrt{3}} \right)$.

Class 12 Physics Wave Optics Brewster's Law and Other methods of polarisation

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The angle of polarisation for a medium with respect to air is $60^{\circ}$. The critical angle of this medium with respect to air is

AP EAMCET 2024 All AP EAMCET

A
$\sin ^{-1} \sqrt{3}$
B
$\tan ^{-1} \sqrt{3}$
C
$\cos ^{-1} \sqrt{3}$
D
$\sin ^{-1} \frac{1}{\sqrt{3}}$

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Wave Optics

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Brewster's Law and Other methods of polarisation

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$A$ ray of light is incident on a transparent glass slab of refractive index $1.62$. The reflected and the refracted rays are mutually perpendicular. The angle of incidence is.......$^o$

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The angle of polarisation for a medium with respect to air is $60^{\circ}$. The critical angle of this medium with respect to air is

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The angle of incidence of light falling on a material is $60^{\circ}$,and the reflected light is completely polarized. The speed of the refracted ray inside the material (in $m/s$) is:

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