A ray of light is incident on a $60^{\circ}$ prism at the minimum deviation position. The angle of refraction

Question

A ray of light is incident on a $60^{\circ}$ prism at the minimum deviation position. The angle of refraction at the first face (i.e., incident face) of the prism is:

Accepted Answer

1.  **Prism Geometry:** In a prism, the angle of the prism ($A$) is related to the angles of refraction at the two faces ($r_1$ and $r_2$) by the equation: $A = r_1 + r_2$.
2.  **Minimum Deviation Condition:** At the position of minimum deviation, the light ray passes symmetrically through the prism. This means the angle of incidence is equal to the angle of emergence ($i = e$), and the angle of refraction at the first face is equal to the angle of incidence at the second face ($r_1 = r_2$).
3.  **Calculation:** Let $r_1 = r_2 = r$. Substituting this into the prism geometry equation gives $A = 2r$, or $r = \frac{A}{2}$.
4.  **Conclusion:** Given that the prism angle $A = 60^{\circ}$, the angle of refraction at the first face is $r = \frac{60^{\circ}}{2} = 30^{\circ}$.

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