The angle of polarisation for a medium is 60^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) According to Brewster's Law,the refractive index $\mu$ of a medium is related to the angle of polarisation $	heta_p$ by the formula: $\mu = 	an

Vedclass · Accepted Answer

$\sin^{-1} \left( \frac{1}{\sqrt{3}} \right)$

Vedclass · Answer

(D) According to Brewster's Law,the refractive index $\mu$ of a medium is related to the angle of polarisation $	heta_p$ by the formula: $\mu = 	an 	heta_p$.Given $	heta_p = 60^o$,we have $\mu = 	an 60^o = \sqrt{3}$.The critical angle $	heta_c$ is related to the refractive index $\mu$ by the formula: $\sin 	heta_c = \frac{1}{\mu}$.Substituting the value of $\mu$,we get $\sin 	heta_c = \frac{1}{\sqrt{3}}$.Therefore,the critical angle is $	heta_c = \sin^{-1} \left( \frac{1}{\sqrt{3}} ight)$.

The angle of polarisation for a medium is $60^o$. What will be the critical angle for this medium?

Similar Questions

In which of the following does the inert pair effect play an important role?

The quantum number which explains the line spectra observed as doublets in the case of hydrogen and alkali metals and doublets and triplets in the case of alkaline earth metals is

If $\sin {\theta _1} + \sin {\theta _2} + \sin {\theta _3} = 3,$ then $\cos {\theta _1} + \cos {\theta _2} + \cos {\theta _3} = $

Let $PS$ be the median of the triangle with vertices $P(2, 2)$,$Q(6, -1)$,and $R(7, 3)$. The equation of the line passing through $(1, -1)$ and parallel to $PS$ is:

In the circuit shown,the heat produced in $5 \Omega$ resistance due to current $I_2$ is $50 \text{ J/s}$. Then,the heat generated per second in $2 \Omega$ resistance is (in $\text{ J/s}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module