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# Light ray is incident on a prism of angle $A = {60^ \circ }$ and refractive index $\mu = \sqrt 2$. The angle of incidence at which the emergent ray grazes the surface is given by:(A) ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)$(B) ${\sin ^{ - 1}}\left( {\dfrac{{1 - \sqrt 3 }}{2}} \right)$(C) ${\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 }}{2}} \right)$(D) ${\sin ^{ - 1}}\left( {\dfrac{2}{{\sqrt 3 }}} \right)$

Last updated date: 14th Apr 2024
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Hint: The angle of incident at which the emergent ray grazes the surface is given by using Snell's law. By using Snell's law, the angle of incident at which the emergent ray grazes the surface can be determined. And by using some trigonometric rules, the solution can be determined.

Complete step by step solution
Given that,
The light ray is incident on a prism of angle $A = {60^ \circ }$,
The refractive index is $\mu = \sqrt 2$.
By Snell’s law,
$\mu \sin {r_2} = 1 \times \sin {90^ \circ }$
The above equation is written as,
$\sin {r_2} = \dfrac{{1 \times \sin {{90}^ \circ }}}{\mu }$
From trigonometry $\sin {90^ \circ } = 1$, then
$\sin {r_2} = \dfrac{1}{\mu }$
By substituting the refractive index value, then
$\sin {r_2} = \dfrac{1}{{\sqrt 2 }}$
By rearranging the terms, then
${r_2} = {\sin ^{ - 1}}\dfrac{1}{{\sqrt 2 }}$
Then the above equation is written as,
${r_2} = {45^ \circ }$
The angle of prism is, ${r_1} + {r_2} = A$
By the above equation, then
${r_1} + {45^ \circ } = {60^ \circ }$
On further simplification, then
${r_1} = {15^ \circ }$
Then by Snell’s law,
$\mu = \dfrac{{\sin i}}{{\sin r}}$
By substituting the refractive index and the angle of refraction in the above equation, then
$\sqrt 2 = \dfrac{{\sin i}}{{\sin {{15}^ \circ }}}$
By rearranging the terms, then the above equation is written as,
$\sin i = \sqrt 2 \sin {15^ \circ }$
Now the above equation is written as,
$\sin i = \sqrt 2 \left[ {\sin \left( {{{45}^ \circ } - {{30}^ \circ }} \right)} \right]$
From the trigonometry, the above equation is written as,
$\sin i = \sqrt 2 \left[ {\sin {{45}^ \circ }\cos {{30}^ \circ } - \cos {{45}^ \circ }\sin {{30}^ \circ }} \right]$
The values of $\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$, $\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$, $\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$ and $\sin {30^ \circ } = \dfrac{1}{2}$, then
$\sin i = \sqrt 2 \left[ {\dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} - \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}} \right]$
By rearranging the terms, then the above equation is written as,
$\sin i = \dfrac{{\sqrt 2 }}{{2\sqrt 2 }}\left[ {\sqrt 3 - 1} \right]$
By cancelling the same terms, then the above equation is written as,
$\sin i = \dfrac{{\sqrt 3 - 1}}{2}$
By rearranging the terms, then the above equation is written as,
$i = {\sin ^{ - 1}}\left( {\dfrac{{\sqrt 3 - 1}}{2}} \right)$
Thus, the above equation shows the angle of incidence at which the emergent ray grazes the surface.

Hence, the option (A) is the correct answer.

Note: By Snell’s law, the refractive index is directly proportional to the sine of the incident angle of the incident ray. And the refractive index is inversely proportional to the sine of the refracted angle of the refracted ray. To solve this problem some trigonometric equations must be known.