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When a glass prism of refracting angle \mathrm{60^{\circ}} is immersed in a liquid its angle of minimum deviation is \mathrm{30^{\circ}}. The critical angle of glass with respect to the liquid medium is:

Option: 1

\mathrm{42^{\circ}}


Option: 2

\mathrm{45^{\circ}}


Option: 3

\mathrm{50^{\circ}}


Option: 4

\mathrm{52^{\circ}}


Answers (1)

best_answer

\mathrm{Here, A=60^{\circ}, \delta_m=30^{\circ}} \\ { }^l \mu_{\mathrm{g}}=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)} \quad{ }^I \mu_{\mathrm{g}}=\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)}=\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\sqrt{2}
Let \mathrm{C} be the critical angle of glass with respect to the liquid medium.
\mathrm{\therefore \quad \sin \mathrm{C}=\frac{1}{{ }^l \mu_{\mathrm{g}}}=\frac{1}{\sqrt{2}} \text { or } \quad \mathrm{C}=\sin ^{-1}\left(\frac{1}{\sqrt{2}}\right)=45^{\circ}}

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Pankaj

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