Question

If the refractive index of material of equilateral prism is $ \sqrt{3} $ then angle of minimum deviation of the prism will be:

• A. $ 25{}^\circ $
• B. $ 60{}^\circ $
• C. $ 45{}^\circ $
• D. $ 30{}^\circ $

Answer 1

Answer: (B)

Refractive index of the material $ \mu =\sqrt{3} $ The relation for refractive index is $ \mu =\frac{\sin i}{\sin r} $ $ \sqrt{3}=\frac{\sin i}{\sin r} $ $ \sqrt{3}=\frac{\sin \frac{A+{{\delta }_{m}}}{2}}{\sin \frac{A}{2}}=\frac{\sin \frac{A+{{\delta }_{m}}}{2}}{\sin \frac{60{}^\circ }{2}} $ $ \upsilon =\frac{\sin \frac{A+{{\delta }_{m}}}{2}}{\frac{1}{2}} $ Or $ \sin \frac{A+{{\delta }_{m}}}{2}=\frac{\sqrt{3}}{2} $ $ =\sin 60{}^\circ $ (where A is the angle of equilateral prism) Hence, $ \frac{A+{{\delta }_{m}}}{2}=60{}^\circ $ Thus, $ A+{{\delta }_{m}}=120 $ $ {{\delta }_{m}}=120{}^\circ -60{}^\circ $ $ {{\delta }_{m}}=60{}^\circ $