Let the speed of light in a medium be V and t | vedclass

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(D) According to Brewster's Law,the refractive index $\mu$ of a medium is given by $\mu = 	an(i_p)$,where $i_p$ is the polarising angle.Also,the refr

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$V \sin(i_p) = C \cos(i_p)$

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(D) According to Brewster's Law,the refractive index $\mu$ of a medium is given by $\mu = 	an(i_p)$,where $i_p$ is the polarising angle.Also,the refractive index $\mu$ is defined as the ratio of the speed of light in vacuum $(C)$ to the speed of light in the medium $(V)$: $\mu = \frac{C}{V}$.Equating the two expressions for $\mu$,we get: $\frac{C}{V} = 	an(i_p)$.This can be written as $\frac{C}{V} = \frac{\sin(i_p)}{\cos(i_p)}$.Rearranging the terms,we get $C \cos(i_p) = V \sin(i_p)$,which is equivalent to $V \sin(i_p) = C \cos(i_p)$.

Let the speed of light in a medium be $V$ and the polarising angle be $i_p$. If $C$ is the speed of light in vacuum,what is the relation between them?

Similar Questions

At what angle of incidence $i$ will the light reflected from water be completely plane-polarized (in $^\circ$)? (Given refractive index of water $\mu = 1.33$)

Light is incident on a glass surface at a polarizing angle of $57.5^{\circ}$. Then the angle between the incident ray and the refracted ray is: (in $^{\circ}$)

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

$A$ beam of light $AO$ is incident on a glass slab $(\mu = 1.54)$ in a direction as shown in the figure. The reflected ray $OB$ is passed through a Nicol prism. On viewing through the rotating Nicol prism,we find that $(\tan 57^{\circ} = 1.54)$.

For a given medium, the polarizing angle is $60^o$. What will be the critical angle for this medium?

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