The angle of minimum deviation for a 75 - prism of dense glass is found to be 45 - when in air and 15 - when immersed in certain liquid- The refractive index of the liquid is -

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

Q. The ratio of angle of minimum deviation produced by a thin prism

(μ = \frac{3}{2})

in air to that in liquid of refractive index

\frac{9}{7}

is :

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The angle of minimum deviation for a 75∘ prism of dense glass is found to be 45∘ when in air and 15∘ when immersed in certain liquid. The refractive index of the liquid is :

The correct option is C √32Weknow,μ12=sinA+D2sinA+D2=sin75+452sin75+152=sin60sin45=√32∴ Option C is correct.

The angle of minimum deviation for a $75^{\circ}$ prism of dense glass is found to be $45^{\circ}$ when in air and $15^{\circ}$ when immersed in certain liquid. The refractive index of the liquid is :

The correct option is C $\sqrt{\frac{3}{2}}$
$\begin{matrix} W e k n o w, μ_{12} = \frac{sin \frac{A + D}{2}}{sin \frac{A + D}{2}} = \frac{sin \frac{75 + 45}{2}}{sin \frac{75 + 15}{2}} = \frac{sin 60}{sin 45} = \sqrt{\frac{3}{2}} \end{matrix}$
$∴$ Option $C$ is correct.