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Derive the relation for the refractive index ($\mu$) of a prism in terms of angle of minimum deviation ($\delta_m$) and angle of prism ($A$).
Derive the relation for the refractive index ($\mu$) of a prism in terms of angle of minimum deviation ($\delta_m$) and angle of prism ($A$).
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At the position of minimum deviation ($\delta = \delta_m$), the refracted ray passing through the interior of the prism becomes perfectly parallel to the base of the prism if the prism is isosceles or equilateral.
Updated On: Jul 2, 2026
- $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
- $\mu = \frac{\cos\left(\frac{A + \delta_m}{2}\right)}{\cos\left(\frac{A}{2}\right)}$
- $\mu = \frac{\sin\left(\frac{A - \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$
- $\mu = \frac{\tan\left(\frac{A + \delta_m}{2}\right)}{\tan\left(\frac{A}{2}\right)}$
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The Correct Option is A
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