JEE Physics 2025 Question with Solution

Given: Prism $P_1$ has angle $A_1 = 4^\circ$ and refractive index $n_1 = 1.54$. Prism $P_2$ has refractive index $n_2 = 1.72$. The combination gives dispersion without deviation.

Find: The angle $A_2$ of prism $P_2$.

For a thin prism, the deviation is

$$\delta = (n-1)A$$

For zero net deviation, the deviations produced by the two prisms must cancel each other. Hence,

$$(n_1-1)A_1 = (n_2-1)A_2$$

Substituting the given values,

$$(1.54-1) \times 4 = (1.72-1) \times A_2$$ $$0.54 \times 4 = 0.72A_2$$ $$2.16 = 0.72A_2$$ $$A_2 = \frac{2.16}{0.72} = 3^\circ$$

Therefore, the angle of prism $P_2$ is $3^\circ$. The correct option is B.

Given: The two thin prisms are arranged so that there is dispersion without deviation.

Find: The magnitude of the angle of prism $P_2$.

Using thin prism deviation,

$$\delta_1 + \delta_2 = 0$$

with

$$\delta_1 = (n_1-1)A_1, \qquad \delta_2 = (n_2-1)A_2$$

So,

$$(n_1-1)A_1 + (n_2-1)A_2 = 0$$

Substituting values,

$$(0.54)(4) + (0.72)A_2 = 0$$ $$2.16 + 0.72A_2 = 0$$ $$0.72A_2 = -2.16$$ $$A_2 = -3^\circ$$

The negative sign only shows that prism $P_2$ is placed inverted relative to prism $P_1$. Hence the required prism angle is the magnitude,

$$|A_2| = 3^\circ$$

Therefore, the correct option remains B.