Minimum Angle of Deviation of a Prism – Rankers Physics
Topic: Ray Optics
Subtopic: Refraction by Prism

Minimum Angle of Deviation of a Prism

Refractive index of a prism is \(cosec(A/2)\). Then minimum angle of deviation is :

\(180^0 –A\)
\(180^0–2A\)
\(90^0–A\)
\(A/2\)

Solution:

Using the prism formula: \(mu = \frac{sin((A+\delta_m)/2)}{sin(A/2)}\). Substituting \(mu = cosec(A/2) =\frac{1}{sin(A/2)}\), we get \(sin((A+\delta_m)/2) = 1 ⇒ (A+\delta_m)/2 = 90^0 ⇒ \delta_m = 180^0 - A\).

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