Monochromatic light is incident on a glass pr | vedclass

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(D) Let the angle of incidence at face $AB$ be $	heta$ and the angle of refraction be $r_1$. At face $AC$,let the angle of incidence be $r_2$ and the

Vedclass · Accepted Answer

$	heta > \sin^{-1}\left[ \mu \sin\left( A - \sin^{-1}\left( \frac{1}{\mu} ight) ight) ight]$

Vedclass · Answer

(D) Let the angle of incidence at face $AB$ be $	heta$ and the angle of refraction be $r_1$. At face $AC$,let the angle of incidence be $r_2$ and the angle of emergence be $e$.For the ray to be transmitted through face $AC$,the angle of incidence $r_2$ must be less than the critical angle $C$,where $\sin C = \frac{1}{\mu}$.Thus,$r_2 From the geometry of the prism,$r_1 + r_2 = A$,so $r_1 = A - r_2$.Since $r_2  A - C$,or $r_1 > A - \sin^{-1}\left( \frac{1}{\mu} ight)$.Applying Snell's law at face $AB$: $\sin 	heta = \mu \sin r_1$.Since the sine function is increasing in the range $[0, \pi/2]$,$r_1 > A - C$ implies $\sin r_1 > \sin(A - C)$.Therefore,$\sin 	heta = \mu \sin r_1 > \mu \sin(A - C)$.Substituting $C = \sin^{-1}(1/\mu)$,we get $\sin 	heta > \mu \sin(A - \sin^{-1}(1/\mu))$.Thus,$	heta > \sin^{-1}\left[ \mu \sin\left( A - \sin^{-1}\left( \frac{1}{\mu} ight) ight) ight]$.

Monochromatic light is incident on a glass prism of angle $A$. If the refractive index of the material of the prism is $\mu$,a ray,incident at an angle $\theta$ on the face $AB$,would get transmitted through the face $AC$ of the prism provided:

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