Question

The refractive index of the material of a prism is √2. If the refracting angle of the prism is 60◦, find:
(1) Angle of minimum deviation:
(2) Angle of incidence:

Answer 1

Angle of Minimum Deviation

Formula for the Refractive Index

The formula for the refractive index \( n \) of the prism is:

\[ n = \frac{\sin \left( A + \delta_m \right)}{2} \div \sin \left( \frac{A}{2} \right) \]

where \( A \) is the angle of the prism and \( \delta_m \) is the angle of minimum deviation.

Given Values

Given \( n = \sqrt{2} \) and \( A = 60^\circ \), we have:

\[ \sqrt{2} = \frac{\sin \left( 60^\circ + \delta_m \right)}{2} \div \sin \left( 30^\circ \right) \]

Simplifying the Equation

Since \( \sin(30^\circ) = \frac{1}{2} \), we can simplify:

\[ \sqrt{2} = 2 \sin \left( 60^\circ + \delta_m \right) \div 2 \]

Simplifying further:

\[ \sqrt{2} = \sin \left( 60^\circ + \delta_m \right) \]

Therefore, we have:

\[ \sin \left( 60^\circ + \delta_m \right) = \sin(45^\circ) \]

Solving for \( \delta_m \), we get:

\[ 60^\circ + \delta_m = 90^\circ \quad \Rightarrow \quad \delta_m = 30^\circ \]

Conclusion

The angle of minimum deviation is \( \delta_m = 30^\circ \).

For the angle of minimum deviation, the angle of incidence equals the angle of emergence. Using symmetry, the angle of incidence \( i_m \) is:

\[ i_m = \frac{A + \delta_m}{2} \]

Substituting the Given Values

Substituting \( A = 60^\circ \) and \( \delta_m = 30^\circ \):

\[ i_m = \frac{60^\circ + 30^\circ}{2} = 45^\circ \]

Conclusion

The angle of incidence is \( i_m = 45^\circ \).

For the angle of minimum deviation, the angle of incidence equals the angle of emergence. Using symmetry, the angle of incidence \( i_m \) is:

\[ i_m = \frac{A + \delta_m}{2} \]

Substituting the Given Values

Substituting \( A = 60^\circ \) and \( \delta_m = 30^\circ \):

\[ i_m = \frac{60^\circ + 30^\circ}{2} = 45^\circ \]

Conclusion

The angle of incidence is \( i_m = 45^\circ \).