Properties of Isosceles Prism at Minimum Deviation Angle

To determine which options are correct, let us analyze the given information step-by-step using the principles of optics.

1. Analysis at Minimum Deviation ($\delta_m = A$)

For a prism of angle $A$, the deviation $\delta$ is given by: $\delta = i_1 + i_2 - A$

At the angle of minimum deviation ($\delta = \delta_m$), the light ray passes symmetrically through the prism: $i_1 = i_2 = i$ $r_1 = r_2 = r = \frac{A}{2}$

Given that $\delta_m = A$: $A = 2i - A \implies 2i = 2A \implies i = A$

Thus, at minimum deviation:

• The angle of incidence at the first surface is $i_1 = A$.
• Since the prism is isosceles and the light passes symmetrically at minimum deviation ($i_1 = i_2$), the refracted ray inside the prism is parallel to the base of the prism.

Therefore, Option (A) is correct.

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2. Relation between Refractive Index $\mu$ and Angle of Prism $A$

Using the prism formula at minimum deviation: $\mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}$

Substituting $\delta_m = A$: $\mu = \frac{\sin\left(\frac{A + A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = \frac{\sin A}{\sin\left(\frac{A}{2}\right)}$

Using the double-angle identity $\sin A = 2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)$: $\mu = \frac{2\sin\left(\frac{A}{2}\right)\cos\left(\frac{A}{2}\right)}{\sin\left(\frac{A}{2}\right)} = 2\cos\left(\frac{A}{2}\right)$

Rearranging this expression for $A$: $\cos\left(\frac{A}{2}\right) = \frac{\mu}{2}$ $\frac{A}{2} = \cos^{-1}\left(\frac{\mu}{2}\right)$ $A = 2\cos^{-1}\left(\frac{\mu}{2}\right)$

Therefore, Option (B) is incorrect.

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3. Relation between $r_1$ and $i_1$ at Minimum Deviation

At minimum deviation, we found: $i_1 = A \quad \text{and} \quad r_1 = \frac{A}{2}$

Substituting $A = i_1$ into the expression for $r_1$: $r_1 = \frac{i_1}{2}$

Therefore, Option (C) is correct.

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4. Condition for Tangential Emergence at the Second Surface

For the emergent ray at the second surface to be tangential, the angle of emergence must be $i_2 = 90^\circ$. By Snell's law at the second surface, the angle of refraction inside the prism at the second surface is equal to the critical angle $\theta_c$: $\sin r_2 = \sin \theta_c = \frac{1}{\mu}$

Since $r_1 + r_2 = A$, we have: $r_1 = A - \theta_c$

Applying Snell's law at the first refracting surface: $\sin i_1 = \mu \sin r_1 = \mu \sin(A - \theta_c)$

Using the sine subtraction formula: $\sin i_1 = \mu \left( \sin A \cos\theta_c - \cos A \sin\theta_c \right)$

Since $\sin\theta_c = \frac{1}{\mu}$, we have $\cos\theta_c = \sqrt{1 - \sin^2\theta_c} = \sqrt{1 - \frac{1}{\mu^2}} = \frac{\sqrt{\mu^2 - 1}}{\mu}$.

Substitute these into the equation for $\sin i_1$: $\sin i_1 = \mu \left( \sin A \frac{\sqrt{\mu^2 - 1}}{\mu} - \cos A \frac{1}{\mu} \right) = \sin A \sqrt{\mu^2 - 1} - \cos A$

From our previous derivation, $\mu = 2\cos\left(\frac{A}{2}\right)$, so: $\mu^2 - 1 = 4\cos^2\left(\frac{A}{2}\right) - 1$

Substituting this back gives: $\sin i_1 = \sin A \sqrt{4\cos^2\left(\frac{A}{2}\right) - 1} - \cos A$ $i_1 = \sin^{-1}\left[\sin A \sqrt{4\cos^2\left(\frac{A}{2}\right) - 1} - \cos A\right]$

Therefore, Option (D) is correct.

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Conclusion

The correct options are A, C, and D.