JEE MAIN - Physics (2015 (Offline) - No. 3)
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 :
_en_3_1.png)
$$\theta > {\cos ^{ - 1}}\left[ {\mu \,\sin \left( {A + {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$$
$$\theta < {\cos ^{ - 1}}\left[ {\mu \,\sin \left( {A + {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$$
$$\theta > si{n^{ - 1}}\left[ {\mu \,\sin \left( {A - {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$$
$$\theta < si{n^{ - 1}}\left[ {\mu \,\sin \left( {A - {{\sin }^{ - 1}}} \right.\left( {{1 \over \mu }} \right)} \right]$$
Explanation
When $${r_2} = C,\,\angle {N_2}Rc = {90^ \circ }$$
Where $$C = $$ critical angle
As $$\sin C = {1 \over v} = \sin {r_2}$$
_en_3_2.png)
Applying snell's law at $$'R'$$
$$\mu \sin {r_2} = 1\sin {90^ \circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
Applying snell's law at $$'Q'$$
$$1 \times \sin \theta = \mu \sin {r_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
But $${r_1} = A - {r_2}$$
So, $$\sin \theta = \mu \sin \left( {A - {r_2}} \right)$$
$$\theta = \mu \sin A\cos {r_2} - \cos \,A\,\,\,\,\,...\left( {iii} \right)$$ [using $$(i)$$]
From $$(1)$$
$$\cos \,{r_2} = \sqrt {1 - {{\sin }^2}{r_2}} = \sqrt {1 - {1 \over {{\mu ^2}}}} \,\,\,\,\,\,\,...\left( {iv} \right)$$
By eq.$$(iii)$$ and $$(iv)$$
$$\sin \theta = \mu \sin A\sqrt {1 - {1 \over {{\mu ^2}}}} - \cos A$$
on further solving we can show for ray not to transmitted through face $$AC$$
$$\theta = {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right)} \right.} \right]$$
So, for transmission through face $$AC$$
$$\theta > {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right.} \right)} \right]$$
Where $$C = $$ critical angle
As $$\sin C = {1 \over v} = \sin {r_2}$$
Applying snell's law at $$'R'$$
$$\mu \sin {r_2} = 1\sin {90^ \circ }\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
Applying snell's law at $$'Q'$$
$$1 \times \sin \theta = \mu \sin {r_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
But $${r_1} = A - {r_2}$$
So, $$\sin \theta = \mu \sin \left( {A - {r_2}} \right)$$
$$\theta = \mu \sin A\cos {r_2} - \cos \,A\,\,\,\,\,...\left( {iii} \right)$$ [using $$(i)$$]
From $$(1)$$
$$\cos \,{r_2} = \sqrt {1 - {{\sin }^2}{r_2}} = \sqrt {1 - {1 \over {{\mu ^2}}}} \,\,\,\,\,\,\,...\left( {iv} \right)$$
By eq.$$(iii)$$ and $$(iv)$$
$$\sin \theta = \mu \sin A\sqrt {1 - {1 \over {{\mu ^2}}}} - \cos A$$
on further solving we can show for ray not to transmitted through face $$AC$$
$$\theta = {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right)} \right.} \right]$$
So, for transmission through face $$AC$$
$$\theta > {\sin ^{ - 1}}\left[ {u\sin \left( {A - {{\sin }^{ - 1}}\left( {{1 \over \mu }} \right.} \right)} \right]$$
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