Question:
When a light ray is incident on a small angle prism of material of refractive index 1.5, the angle of minimum deviation is $7^\circ$. If the prism is immersed in a liquid of refractive index 1.2, then the angle of minimum deviation is:
When a light ray is incident on a small angle prism of material of refractive index 1.5, the angle of minimum deviation is $7^\circ$. If the prism is immersed in a liquid of refractive index 1.2, then the angle of minimum deviation is:
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For a given small-angle prism, deviation is directly proportional to $(\mu_{\text{rel}} - 1)$.
Ratio: $\frac{\delta_2}{\delta_1} = \frac{\mu_{\text{rel}} - 1}{\mu_g - 1} = \frac{1.25 - 1}{1.5 - 1} = \frac{0.25}{0.5} = \frac{1}{2}$.
Thus, $\delta_2 = \frac{1}{2} \delta_1 = \frac{7^\circ}{2} = 3.5^\circ$.
This direct ratio method is extremely fast.
Ratio: $\frac{\delta_2}{\delta_1} = \frac{\mu_{\text{rel}} - 1}{\mu_g - 1} = \frac{1.25 - 1}{1.5 - 1} = \frac{0.25}{0.5} = \frac{1}{2}$.
Thus, $\delta_2 = \frac{1}{2} \delta_1 = \frac{7^\circ}{2} = 3.5^\circ$.
This direct ratio method is extremely fast.
Updated On: Jul 22, 2026
- $10.5^\circ$
- $1.75^\circ$
- $3.5^\circ$
- $14^\circ$
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We need to find how the angle of minimum deviation of a small-angle prism changes when it is moved from air into a liquid of a given refractive index.
Step 2: Key Formula and Approach:
For a prism with a small refracting angle $A$, the angle of minimum deviation $\delta$ is:
\[ \delta = (\mu_{\text{rel}} - 1) A \] where $\mu_{\text{rel}}$ is the refractive index of the prism relative to the surrounding medium.
We will find $A$ using the first case (in air) and then calculate $\delta$ in the second case (in the liquid).
Step 3: Detailed Explanation:
• Case 1: Prism in air ($\mu_{\text{air}} = 1$):
Refractive index of prism: $\mu_g = 1.5$
Minimum deviation: $\delta_1 = 7^\circ$
Using the small angle formula:
\[ \delta_1 = (\mu_g - 1) A \] \[ 7^\circ = (1.5 - 1) A = 0.5 A \] \[ A = \frac{7^\circ}{0.5} = 14^\circ \quad \text{--- (Refracting angle of prism)} \]
• Case 2: Prism in liquid ($\mu_{\text{liq}} = 1.2$):
The relative refractive index of the prism is now:
\[ \mu_{\text{rel}} = \frac{\mu_g}{\mu_{\text{liq}}} = \frac{1.5}{1.2} = 1.25 \] Calculate the new angle of minimum deviation $\delta_2$:
\[ \delta_2 = (\mu_{\text{rel}} - 1) A \] \[ \delta_2 = (1.25 - 1) \times 14^\circ \] \[ \delta_2 = 0.25 \times 14^\circ = \frac{14^\circ}{4} = 3.5^\circ \]
Step 4: Final Answer:
The angle of minimum deviation when immersed in the liquid is $3.5^\circ$, which corresponds to Option (C).
We need to find how the angle of minimum deviation of a small-angle prism changes when it is moved from air into a liquid of a given refractive index.
Step 2: Key Formula and Approach:
For a prism with a small refracting angle $A$, the angle of minimum deviation $\delta$ is:
\[ \delta = (\mu_{\text{rel}} - 1) A \] where $\mu_{\text{rel}}$ is the refractive index of the prism relative to the surrounding medium.
We will find $A$ using the first case (in air) and then calculate $\delta$ in the second case (in the liquid).
Step 3: Detailed Explanation:
• Case 1: Prism in air ($\mu_{\text{air}} = 1$):
Refractive index of prism: $\mu_g = 1.5$
Minimum deviation: $\delta_1 = 7^\circ$
Using the small angle formula:
\[ \delta_1 = (\mu_g - 1) A \] \[ 7^\circ = (1.5 - 1) A = 0.5 A \] \[ A = \frac{7^\circ}{0.5} = 14^\circ \quad \text{--- (Refracting angle of prism)} \]
• Case 2: Prism in liquid ($\mu_{\text{liq}} = 1.2$):
The relative refractive index of the prism is now:
\[ \mu_{\text{rel}} = \frac{\mu_g}{\mu_{\text{liq}}} = \frac{1.5}{1.2} = 1.25 \] Calculate the new angle of minimum deviation $\delta_2$:
\[ \delta_2 = (\mu_{\text{rel}} - 1) A \] \[ \delta_2 = (1.25 - 1) \times 14^\circ \] \[ \delta_2 = 0.25 \times 14^\circ = \frac{14^\circ}{4} = 3.5^\circ \]
Step 4: Final Answer:
The angle of minimum deviation when immersed in the liquid is $3.5^\circ$, which corresponds to Option (C).
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