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When a prism is in minimum deviation position, the value of angle of incidence is equal to:

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Question

When a prism is in minimum deviation position, the value of angle of incidence is equal to:

Options

  • \(\dfrac{\angle \mathrm{A} + \angle \delta}{2}\)

  • \((\angle \mathrm{A}+\angle \delta) \times 2\)

  • \(\dfrac{2}{\angle \mathrm{A} + \angle \delta}\)

  • none of these

MCQ
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Solution

\(\dfrac{\angle \mathrm{A} + \angle \delta}{2}\)

Explanation:

At minimum deviation, the angle of incidence equals the angle of emergence.

Using the prism relation $$\angle \mathrm{A} + \angle \delta = \angle i + \angle e$$, we get $$\angle i = \dfrac{\angle \mathrm{A} + \angle \delta}{2}.$$

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Chapter 4: Refraction of Light - EXERCISE-2 [Page 85]

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Goyal Brothers Prakashan A New Approach to ICSE Physics [English] Class 10
Chapter 4 Refraction of Light
EXERCISE-2 | Q 18. | Page 85
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