Question 242: Ray Optics

Question

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?

Accepted Answer

We know that $$i + e - A = \delta $$
$${35^ \circ } + {79^ \circ } - A = {40^ \circ }$$
$$	herefore$$ $$A = {74^ \circ }$$
But $$\mu  = {{\sin \left( {{{A + {\delta _m}} \over 2}} ight)} \over {\sin A/2}} = {{\sin \left( {{{74 + \delta } \over 2}} ight)} \over {\sin {{74} \over 2}}}$$
$$ = {5 \over 3}\sin \left( {{{37}^ \circ } + {{{\delta _m}} \over 2}} ight)$$
$${\mu _{\max }}\,$$ 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
$$\mu  < {5 \over 3}\sin \,{57^ \circ } < {5 \over 3}\,\,$$ $$\sin {60^ \circ } \Rightarrow \mu  = 1.45$$

Answer

$$1.7$$

Answer

$$1.8$$

Answer

$$1.5$$

Answer

$$1.6$$

Ray Optics

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