Q130 / 150 · MHT CET 2026 (11 April Shift 2) · Physics
A thin prism in air produces the angle of minimum deviation $\delta$. If the prism is immersed in water, the angle of minimum deviation for the same ray is (refractive index of water is $4/3$ and that of glass prism is $3/2$)
- A$\delta$
- B$\dfrac{\delta}{2}$
- Correct answerC$\dfrac{\delta}{4}$
- D$2\delta$
Solution
The deviation produced by a thin prism in air is given by $\delta = (\mu_g - 1)A$.
Substituting $\mu_g = \dfrac{3}{2}$, we get $\delta = \left(\dfrac{3}{2} - 1\right)A = \dfrac{A}{2}$.
When the prism is immersed in water, the relative refractive index of glass with respect to water is $\mu_{gw} = \dfrac{\mu_g}{\mu_w} = \dfrac{3/2}{4/3} = \dfrac{9}{8}$.
The new angle of minimum deviation is $\delta' = (\mu_{gw} - 1)A$.
Substituting $\mu_{gw} = \dfrac{9}{8}$, we get $\delta' = \left(\dfrac{9}{8} - 1\right)A = \dfrac{A}{8}$.
Comparing the two deviations, $\dfrac{\delta'}{\delta} = \dfrac{A/8}{A/2} = \dfrac{1}{4}$.
Thus, $\delta' = \dfrac{\delta}{4}$.
Answer: $\dfrac{\delta}{4}$
Substituting $\mu_g = \dfrac{3}{2}$, we get $\delta = \left(\dfrac{3}{2} - 1\right)A = \dfrac{A}{2}$.
When the prism is immersed in water, the relative refractive index of glass with respect to water is $\mu_{gw} = \dfrac{\mu_g}{\mu_w} = \dfrac{3/2}{4/3} = \dfrac{9}{8}$.
The new angle of minimum deviation is $\delta' = (\mu_{gw} - 1)A$.
Substituting $\mu_{gw} = \dfrac{9}{8}$, we get $\delta' = \left(\dfrac{9}{8} - 1\right)A = \dfrac{A}{8}$.
Comparing the two deviations, $\dfrac{\delta'}{\delta} = \dfrac{A/8}{A/2} = \dfrac{1}{4}$.
Thus, $\delta' = \dfrac{\delta}{4}$.
Answer: $\dfrac{\delta}{4}$
Hence, the correct answer is option: 'C'.