Question

The refractive index for material of prism having refracting angle $60^\circ$ is $1.414$. For light incident on prism in minimum deviation position, calculate:
(i) Angle of minimum deviation,
(ii) Angle of incidence,
(iii) Angle of refraction,
(iv) Angle of emergence. Also draw ray diagram.

Hint:

In minimum deviation, the path of light through prism is symmetric: $i = e$, $r_1 = r_2 = A/2$.

Answer 1

Step 1: Formula for refractive index of prism.
\[ \mu = \frac{\sin \left(\tfrac{A + D_m}{2}\right)}{\sin \left(\tfrac{A}{2}\right)} \] where $A$ = refracting angle, $D_m$ = angle of minimum deviation.
Step 2: Substitution.
\[ 1.414 = \frac{\sin \left(\tfrac{60 + D_m}{2}\right)}{\sin 30^\circ} \] Since $\sin 30^\circ = 0.5$: \[ 1.414 \times 0.5 = \sin \left(\tfrac{60 + D_m}{2}\right) \] \[ 0.707 = \sin \left(\tfrac{60 + D_m}{2}\right) \]
Step 3: Solve for $D_m$.
\[ \frac{60 + D_m}{2} = 45^\circ \quad \Rightarrow \quad D_m = 30^\circ \]
Step 4: Angle of incidence at minimum deviation.
At minimum deviation: \[ i = e = \frac{A + D_m}{2} = \frac{60 + 30}{2} = 45^\circ \]
Step 5: Angle of refraction inside prism.
\[ r = \frac{A}{2} = \frac{60}{2} = 30^\circ \]
Step 6: Angle of emergence.
At symmetry, $e = i = 45^\circ$.
Step 7: Conclusion.
- $D_m = 30^\circ$
- $i = 45^\circ$
- $r = 30^\circ$
- $e = 45^\circ$