The angle of incidence for a ray of light at a refracting surface of a prism is $45^\circ$. The angle of prism

Question

The angle of incidence for a ray of light at a refracting surface of a prism is $45^\circ$. The angle of prism is $60^\circ$. If the ray suffers minimum deviation through the prism, the angle of deviation and refractive index of the material of the prism respectively are:

Accepted Answer

1.  **Condition for Minimum Deviation:** When a prism is in the position of minimum deviation, the angle of incidence ($i$) is equal to the angle of emergence ($e$), i.e., $i = e$. Also, the angle of refraction inside the prism at both surfaces is the same ($r_1 = r_2 = r = A/2$).
2.  **Calculate Angle of Deviation ($\delta_m$):**
    *   The relationship between the prism angle ($A$), deviation ($\delta$), incidence ($i$), and emergence ($e$) is given by $A + \delta = i + e$.
    *   At minimum deviation: $A + \delta_m = 2i$.
    *   Given: $i = 45^\circ$ and $A = 60^\circ$.
    *   $60^\circ + \delta_m = 2(45^\circ) = 90^\circ$.
    *   $\delta_m = 90^\circ - 60^\circ = 30^\circ$.
3.  **Calculate Refractive Index ($\mu$):**
    *   Using the prism formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(A/2)}$.
    *   Substituting values: $\mu = \frac{\sin(\frac{60^\circ + 30^\circ}{2})}{\sin(60^\circ/2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)}$.
    *   $\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$.
4.  **Conclusion:** The angle of deviation is $30^\circ$ and the refractive index is $\sqrt{2}$.

Step-by-Step Solution

Practice All PHYSICS Questions