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  • For a prism kept in air it is found that for an angle of incidence  , the angle of prism '

For a prism kept in air it is found that for an angle of incidence 60^{\circ} , the angle of prism ' A ', angle of deviation '\delta  ' and angle of emergence 'e' become equal. Then the refractive index of the prism is

Option: 1

1.73


Option: 2

1.15


Option: 3

1.5


Option: 4

1.33


Answers (1)

best_answer

Given \mathrm{\mathrm{i}=60^{\circ}~ and ~\mathrm{A}=\delta=\mathrm{e} }

\mathrm{ \delta=\mathrm{i}+\mathrm{e}-\mathrm{A} \Rightarrow \delta=\mathrm{i}(\because \mathrm{e}=\mathrm{A}) \\ }
\mathrm{ \mu=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \frac{\mathrm{A}}{2}} }  Here angle of deviation \mathrm{ (\delta) }  is minimum \mathrm{ (\because \mathrm{i}=\mathrm{e}) \\ }
\mathrm{ \mu=\frac{\sin \left(\frac{60^{\circ}+60^{\circ}}{2}\right)}{\sin \left(60^{\circ} / 2\right)}=1.732 }.
 

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Divya Prakash Singh

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