Question:
The refracting angle of a prism is \(A\) and the refractive index of the material of the prism is \(\cot\left(\dfrac{A}{2}
ight)\). Then the angle of minimum deviation will be
The refracting angle of a prism is \(A\) and the refractive index of the material of the prism is \(\cot\left(\dfrac{A}{2}
ight)\). Then the angle of minimum deviation will be
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Use the prism formula \(\mu=\dfrac{\sin rac{A+\delta_m}{2}}{\sin rac{A}{2}}\) and \(\cot rac{A}{2}=\sin(90^\circ- rac{A}{2})/\sin rac{A}{2}\).
Updated On: Nov 4, 2026
- \(180^\circ - 2A\)
- \(90^\circ - A\)
- \(180^\circ + 2A\)
- \(180^\circ - 3A\)
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The Correct Option is A
Solution and Explanation
For a prism, \(\mu=\dfrac{\sin\left( rac{A+\delta_m}{2}
ight)}{\sin\left( rac{A}{2}
ight)}\).
Given \(\mu=\cot\dfrac{A}{2}=\dfrac{\cos rac{A}{2}}{\sin rac{A}{2}}\).
Hence \(\sin\left(\dfrac{A+\delta_m}{2} ight)=\cos\dfrac{A}{2}=\sin\left(90^\circ-\dfrac{A}{2} ight)\).
So \(\dfrac{A+\delta_m}{2}=90^\circ-\dfrac{A}{2}\) which gives \(A+\delta_m=180^\circ-A\).
Therefore \(\delta_m=180^\circ-2A\).
Answer: option (1)
Given \(\mu=\cot\dfrac{A}{2}=\dfrac{\cos rac{A}{2}}{\sin rac{A}{2}}\).
Hence \(\sin\left(\dfrac{A+\delta_m}{2} ight)=\cos\dfrac{A}{2}=\sin\left(90^\circ-\dfrac{A}{2} ight)\).
So \(\dfrac{A+\delta_m}{2}=90^\circ-\dfrac{A}{2}\) which gives \(A+\delta_m=180^\circ-A\).
Therefore \(\delta_m=180^\circ-2A\).
Answer: option (1)
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