JEE MAIN - Physics (2016 (Offline) - No. 9)
In an experiment for determination of refractive index of glass of a prism by $$i - \delta ,$$ plot it was found thata ray incident at angle $${35^ \circ }$$, suffers a deviation of $${40^ \circ }$$ and that it emerges at angle $${79^ \circ }.$$ In that case which of the following is closest to the maximum possible value of the refractive index?
$$1.7$$
$$1.8$$
$$1.5$$
$$1.6$$
Explanation
We know that $$i + e - A = \delta $$
$${35^ \circ } + {79^ \circ } - A = {40^ \circ }$$
$$\therefore$$ $$A = {74^ \circ }$$
But $$\mu = {{\sin \left( {{{A + {\delta _m}} \over 2}} \right)} \over {\sin A/2}} = {{\sin \left( {{{74 + \delta } \over 2}} \right)} \over {\sin {{74} \over 2}}}$$
$$ = {5 \over 3}\sin \left( {{{37}^ \circ } + {{{\delta _m}} \over 2}} \right)$$
$${\mu _{\max }}\,$$ can be $${5 \over 3}.$$ That is $${\mu _{\max }}$$ is less than $${5 \over 3} = 1.67$$
But $${\delta _m}$$ will be less than $${40^ \circ }$$ so
$$\mu < {5 \over 3}\sin \,{57^ \circ } < {5 \over 3}\,\,$$ $$\sin {60^ \circ } \Rightarrow \mu = 1.45$$
$${35^ \circ } + {79^ \circ } - A = {40^ \circ }$$
$$\therefore$$ $$A = {74^ \circ }$$
But $$\mu = {{\sin \left( {{{A + {\delta _m}} \over 2}} \right)} \over {\sin A/2}} = {{\sin \left( {{{74 + \delta } \over 2}} \right)} \over {\sin {{74} \over 2}}}$$
$$ = {5 \over 3}\sin \left( {{{37}^ \circ } + {{{\delta _m}} \over 2}} \right)$$
$${\mu _{\max }}\,$$ can be $${5 \over 3}.$$ That is $${\mu _{\max }}$$ is less than $${5 \over 3} = 1.67$$
But $${\delta _m}$$ will be less than $${40^ \circ }$$ so
$$\mu < {5 \over 3}\sin \,{57^ \circ } < {5 \over 3}\,\,$$ $$\sin {60^ \circ } \Rightarrow \mu = 1.45$$
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