The refractive index (R.I.) of a prism is sqr | vedclass

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(A) For a ray to be transmitted through the prism,the angle of refraction at the second surface $(r_2)$ must be less than or equal to the critical ang

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(A) For a ray to be transmitted through the prism,the angle of refraction at the second surface $(r_2)$ must be less than or equal to the critical angle $(	heta_C)$.$	heta_C = \sin^{-1}\left(\frac{1}{\mu}ight) = \sin^{-1}\left(\sqrt{\frac{3}{7}}ight)$.Given $A = r_1 + r_2$,for the limiting case,we set $r_2 = 	heta_C$.Thus,$\sin r_2 = \sqrt{\frac{3}{7}}$.Using Snell's law at the first surface: $\mu = \frac{\sin i_1}{\sin r_1}$,where $r_1 = A - r_2$.$\sin r_1 = \sin(60^{\circ} - r_2) = \sin 60^{\circ} \cos r_2 - \cos 60^{\circ} \sin r_2$.Since $\sin r_2 = \sqrt{\frac{3}{7}}$,then $\cos r_2 = \sqrt{1 - \frac{3}{7}} = \sqrt{\frac{4}{7}} = \frac{2}{\sqrt{7}}$.$\sin r_1 = \left(\frac{\sqrt{3}}{2} 	imes \frac{2}{\sqrt{7}}ight) - \left(\frac{1}{2} 	imes \sqrt{\frac{3}{7}}ight) = \frac{\sqrt{3}}{\sqrt{7}} - \frac{\sqrt{3}}{2\sqrt{7}} = \frac{\sqrt{3}}{2\sqrt{7}}$.Now,$\sin i_1 = \mu \sin r_1 = \sqrt{\frac{7}{3}} 	imes \frac{\sqrt{3}}{2\sqrt{7}} = \frac{1}{2}$.Therefore,$i_1 = \sin^{-1}(0.5) = 30^{\circ}$.

The refractive index $(R.I.)$ of a prism is $\sqrt{\frac{7}{3}}$ and the angle of the prism is $60^{\circ}$. The limiting angle of incidence of a ray that will be transmitted through the prism is .....$^{\circ}$

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