A light ray is incident at an angle of 60^o o | vedclass

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(D) Given: Angle of incidence $i = 60^o$,Prism angle $A = 45^o$,Emergent angle $e = 90^o$ (relative to the surface,so the angle of emergence $e' = 0^o

Vedclass · Accepted Answer

$\mu = \sqrt{\frac{3}{2}}, \delta = 15^o$

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(D) Given: Angle of incidence $i = 60^o$,Prism angle $A = 45^o$,Emergent angle $e = 90^o$ (relative to the surface,so the angle of emergence $e' = 0^o$ relative to the normal).$1$. From the geometry of the prism,the angle of refraction at the second surface $r_2 = 0^o$ because the ray emerges normally.$2$. Using the relation $A = r_1 + r_2$,we get $45^o = r_1 + 0^o$,so $r_1 = 45^o$.$3$. Using Snell's law at the first surface: $\mu = \frac{\sin i}{\sin r_1} = \frac{\sin 60^o}{\sin 45^o} = \frac{\sqrt{3}/2}{1/\sqrt{2}} = \sqrt{\frac{3}{2}}$.$4$. The angle of deviation $\delta$ is given by $\delta = i + e' - A$. Since the ray emerges normally,the angle of emergence relative to the normal is $e' = 0^o$.$5$. Thus,$\delta = 60^o + 0^o - 45^o = 15^o$.Therefore,$\mu = \sqrt{\frac{3}{2}}$ and $\delta = 15^o$.

$A$ light ray is incident at an angle of $60^o$ on a prism of angle $45^o$. When the light ray emerges from the other surface at $90^o$ to the surface,the refractive index of the material of the prism $\mu$ and the angle of deviation $\delta$ are given by:

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