Question:
When a light ray incidents on an equilateral prism of material of refractive index $\sqrt{2}$, the angle of minimum deviation is D. If the light ray incidents on another equilateral prism of material of refractive index $\sqrt{3}$, then the angle of minimum deviation is:
When a light ray incidents on an equilateral prism of material of refractive index $\sqrt{2}$, the angle of minimum deviation is D. If the light ray incidents on another equilateral prism of material of refractive index $\sqrt{3}$, then the angle of minimum deviation is:
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$\sin(A+D)/2 = \mu \sin(A/2)$.
Updated On: Jun 6, 2026
- $\sqrt{1.5} D$
- $\sqrt{3} D$
- 0.5 D
- 2 D
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The Correct Option is D
Solution and Explanation
Step 1: Concept
Minimum deviation formula: $\mu = \frac{\sin((A+D)/2)}{\sin(A/2)}$.
Step 2: Meaning
Prism angle $A=60^{\circ}$. $\mu_1 = \sqrt{2}$, $\mu_2 = \sqrt{3}$.
Step 3: Analysis
For $\mu_1=\sqrt{2}$: $\sqrt{2} = \frac{\sin(30+D/2)}{\sin(30)} \rightarrow \sin(30+D/2) = \sqrt{2} \cdot 0.5 = 0.707 \rightarrow 30+D/2 = 45 \rightarrow D=30^{\circ}$. For $\mu_2=\sqrt{3}$: $\sqrt{3} = \frac{\sin(30+D'/2)}{0.5} \rightarrow \sin(30+D'/2) = \sqrt{3}/2 \rightarrow 30+D'/2 = 60 \rightarrow D'=60^{\circ}$. $D'=2D$.
Step 4: Conclusion
The new deviation is 2D.
Final Answer: (D)
Minimum deviation formula: $\mu = \frac{\sin((A+D)/2)}{\sin(A/2)}$.
Step 2: Meaning
Prism angle $A=60^{\circ}$. $\mu_1 = \sqrt{2}$, $\mu_2 = \sqrt{3}$.
Step 3: Analysis
For $\mu_1=\sqrt{2}$: $\sqrt{2} = \frac{\sin(30+D/2)}{\sin(30)} \rightarrow \sin(30+D/2) = \sqrt{2} \cdot 0.5 = 0.707 \rightarrow 30+D/2 = 45 \rightarrow D=30^{\circ}$. For $\mu_2=\sqrt{3}$: $\sqrt{3} = \frac{\sin(30+D'/2)}{0.5} \rightarrow \sin(30+D'/2) = \sqrt{3}/2 \rightarrow 30+D'/2 = 60 \rightarrow D'=60^{\circ}$. $D'=2D$.
Step 4: Conclusion
The new deviation is 2D.
Final Answer: (D)
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