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Q.
A prism is filled with liquid of refractive index of $ \sqrt{2} $ If angle of prism is $ 60{}^\circ $ , find angle of minimum deviation
VMMC MedicalVMMC Medical 2002
Solution:
Applying the formula for refractive index $ (\mu ) $ of a prism having angle of prism A $ \mu =\frac{\sin \frac{A+\delta m}{2}}{\sin \frac{A}{2}} $ ?(1) Putting the given values of $ \mu =\sqrt{2} $ and angle of prism $ \text{A = 6}{{\text{0}}^{\text{o}}} $ in the Eq. (1), we get $ \sqrt{2}=\frac{\sin \frac{{{60}^{o}}+\delta m}{2}}{\sin \frac{{{60}^{o}}}{2}}=\frac{\sin \frac{{{60}^{o}}+\delta m}{2}}{\sin {{30}^{o}}} $ or $ \sin \left( \frac{{{60}^{o}}+\delta m}{2} \right)=\sqrt{2}\times \frac{1}{2}=\frac{1}{\sqrt{2}} $ or $ \sin \frac{{{60}^{o}}+\delta m}{2}=\sin {{45}^{o}} $ or $ {{60}^{o}}+\delta m={{90}^{o}} $ or $ {{\delta }_{m}}={{90}^{o}}-{{60}^{o}}={{30}^{o}} $