A large glass slab (mu = 5/3) of thickness 8 | vedclass

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(B) The light rays from the point source will emerge from the top surface only if the angle of incidence at the top surface is less than or equal to t

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The light rays from the point source will emerge from the top surface only if the angle of incidence at the top surface is less than or equal to the critical angle $	heta_c$.The critical angle $	heta_c$ is given by $\sin 	heta_c = 1/\mu$.Given $\mu = 5/3$,we have $\sin 	heta_c = 1 / (5/3) = 3/5$.From the geometry of the problem,$	an 	heta_c = R / h$,where $h = 8 	ext{ cm}$ is the thickness of the slab.Since $\sin 	heta_c = 3/5$,we can form a right-angled triangle with opposite side $3$ and hypotenuse $5$. The adjacent side is $\sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4$.Thus,$	an 	heta_c = 3/4$.Substituting the values,$3/4 = R / 8$.$R = (3 	imes 8) / 4 = 6 	ext{ cm}$.Therefore,the value of $R$ is $6$.

$A$ large glass slab $(\mu = 5/3)$ of thickness $8 \text{ cm}$ is placed over a point source of light on a plane surface. It is observed that light emerges from the top surface of the slab through a circular area of radius $R \text{ cm}$. What is the value of $R$?

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