Question

The ratio of the angle of deviation produced by a thin prism, when it is placed in air to the angle of deviation produced when it is immersed in water of refractive index $\frac{4}{3}$ is: (Take refractive index of glass $= \frac{3}{2}$)}

• A. $1:4$
• B. $9:8$
• C. $4:1$
• D. $8:9$

Hint:

For a standard glass prism ($\mu = 1.5$) dipped in water ($\mu = 1.33$), the angle of deviation always drops to exactly one-fourth of its original value in air ($\delta_{\text{water}} = \frac{\delta_{\text{air}}}{4}$). Remembering this common factor saves calculation steps!

Answer 1

The Correct Option is C

Concept: For a thin prism of refracting angle $A$, the angle of minimum deviation ($\delta$) depends on the relative refractive index of the prism material with respect to its surrounding medium ($\mu_{\text{relative}}$): \[ \delta = (\mu_{\text{relative}} - 1)A \] When placed in air, $\mu_{\text{relative}} = \mu_g$. When immersed in a liquid, $\mu_{\text{relative}} = \frac{\mu_g}{\mu_l}$.

Step 1: Calculate deviation in air ($\delta_{\text{air}}$).
Given the refractive index of glass $\mu_g = \frac{3}{2}$: \[ \delta_{\text{air}} = (\mu_g - 1)A = \left(\frac{3}{2} - 1\right)A = \frac{1}{2}A \]

Step 2: Calculate deviation in water ($\delta_{\text{water}}$).
Given the refractive index of water $\mu_w = \frac{4}{3}$: \[ \delta_{\text{water}} = \left(\frac{\mu_g}{\mu_w} - 1\right)A = \left(\frac{3/2}{4/3} - 1\right)A = \left(\frac{9}{8} - 1\right)A = \frac{1}{8}A \]

Step 3: Compute the ratio of the two deviations.
\[ \text{Ratio} = \frac{\delta_{\text{air}}}{\delta_{\text{water}}} = \frac{\frac{1}{2}A}{\frac{1}{8}A} = \frac{8}{2} = \frac{4}{1} \implies 4:1 \]