CBSE 2025, Set 2

Question

A ray of light is incident on face AB of a prism ABC with angle of prism $A$ and emerges out from face AC. The prism is set in the position of minimum deviation with angle of deviation $\delta$. Find (i) the angle of incidence and (ii) the angle of refraction on face AB.

Edvaya · Accepted Answer

The correct answer is Subjective. For refraction through a prism, the standard relations are:
$$A + \delta = i + e$$
$$A = r_1 + r_2$$
where $A$ is the prism angle, $\delta$ is the angle of deviation, $i$ is the angle of incidence, $e$ is the angle of emergence, and $r_1, r_2$ are the angles of refraction at the two faces.

At the position of **minimum deviation** ($\delta = \delta_{min}$):
* The ray passes symmetrically through the prism, so the angle of incidence equals the angle of emergence: $i = e$.
* The angles of refraction at the two faces are equal: $r_1 = r_2 = r$.

**(i) Angle of incidence ($i$):**
Substitute $i = e$ into the deviation equation:
$$A + \delta = i + i$$
$$A + \delta = 2i$$
$$\boxed{i = \frac{A + \delta}{2}}$$

**(ii) Angle of refraction on face AB ($r$):**
Substitute $r_1 = r_2 = r$ into the prism angle equation:
$$A = r + r$$
$$A = 2r$$
$$\boxed{r = \frac{A}{2}}$$

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