Question

A ray of light is incident at angle of 60 on root3 cm thick plate the shift in the path of the ray as it emerges out from the plate is

Answer 1

Answer:

Sorry for bad hand writing

Answer 2

Answer:

The shift in the path of the ray as it emerges out from the plate is 1cm.

Explanation:

Given angle of incidence of ray of light, $i=60^{o}$

Thickness of plate, $t=\sqrt{3}~cm$

As it is not given in the question,

but assuming that the refractive index should be, $\mu=\sqrt3}$

Using Snell's law, $\mu=\frac{sin~i}{sin~r}$

$\implies {sin~r}=\frac{sin~i}{{\mu}}=\frac{sin~60^{o} }{{\sqrt{3} }}$

$=\frac{\frac{\sqrt{3} }{2} }{{\sqrt{3} }}=\frac{1}{2}$

${sin~r}=\frac{1}{2} \implies r=30^{o}$

Therefore, angle of refraction, $r=30^{o}$

The lateral shift of the emergent ray is given by

$x=t\times \frac{sin(i-r)}{cos~r}$

$=\sqrt{3} \times \frac{sin(60-30)}{cos~30}$

$=\sqrt{3} \times \frac{sin(30)}{cos~30}=\sqrt{3} \times\frac{\frac{1}{2} }{\frac{\sqrt{3} }{2} }$

$=\sqrt{3} \times {\frac{1}{2} }\times {\frac{ 2}{\sqrt{3}} } =1$

Hence,  the shift in the path of the ray as it emerges out from the plate is 1cm.