For an isosceles prism of angle A and refractive index μ, it is found that the angle of minimum devi

Q: Why is $r_1 = r_2 = A/2$ at minimum deviation?

At the angle of minimum deviation, a light ray passes symmetrically through the prism. Because of this symmetry, the angle of incidence equals the angle of emergence ($i_1 = i_2$), and consequently, the angles of refraction inside the prism are also equal ($r_1 = r_2$). Since $r_1 + r_2 = A$, it follows that $2r_1 = A$, or $r_1 = A/2$.

Q: How did we get $\mu = 2 \cos(A/2)$ from $\sin A = \mu \sin(A/2)$?

We used the double-angle trigonometric identity for sine, which states that $\sin(2\theta) = 2 \sin \theta \cos \theta$. By treating $A$ as $2 \times (A/2)$, we can write $\sin A = 2 \sin(A/2) \cos(A/2)$. Substituting this into the equation allows us to cancel the $\sin(A/2)$ term on both sides, leaving $\mu = 2 \cos(A/2)$.

Q: What exactly does 'tangential to the surface' mean in option C?

When a ray emerges 'tangential to the surface', it means it grazes along the boundary of the prism as it exits. In optical terms, this is called grazing emergence, and it corresponds to an angle of emergence $i_2$ of exactly $90^\circ$.

Q: Why is $\cos \theta_c = \frac{\sqrt{\mu^2 - 1}}{\mu}$?

By definition, the critical angle $\theta_c$ is given by $\sin \theta_c = \frac{1}{\mu}$. Using the fundamental trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$, we can solve for cosine: $\cos \theta_c = \sqrt{1 - \sin^2 \theta_c} = \sqrt{1 - (1/\mu)^2} = \sqrt{\frac{\mu^2 - 1}{\mu^2}} = \frac{\sqrt{\mu^2 - 1}}{\mu}$.

Q: Is the ray inside an isosceles prism always parallel to the base at minimum deviation?

Not necessarily! It is only parallel to the base if the two angles adjacent to the base are equal (i.e., it's a symmetric isosceles prism). If the prism angle $A$ is one of the equal angles, the ray won't be parallel to the base. However, in JEE problems, 'isosceles prism' usually implies the symmetric case, making option A correct under that assumption.

Question

For an isosceles prism of angle $A$ and refractive index $\mu$, it is found that the angle of minimum deviation $\delta_m = A$. Which of the following options is/are correct?

Options:

(A) For the angle of incidence $i_1 = A$, the ray inside the prism is parallel to the base of the prism

(B) At minimum deviation, the incident angle $i_1$ and the refracting angle $r_1$ at the first refracting surface are related by $r_1 = \left(\frac{i_1}{2}\right)$

(C) For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is $i_1 = \sin^{-1}\left[\sin A \sqrt{4\cos^2 \frac{A}{2} - 1} - \cos A\right]$

(D) For this prism, the refractive index $\mu$ and the angle prism $A$ are related as $A = \frac{1}{2}\cos^{-1}\left(\frac{\mu}{2}\right)$

Sigma Percentile · Accepted Answer

The beauty of optics often lies in the elegant symmetry of light rays traversing through geometric shapes. In this thrilling problem from JEE Advanced 2017, we are tasked with analyzing an isosceles prism under two distinct and fascinating conditions: minimum deviation and grazing emergence.

### The Symmetry of Minimum Deviation

We are given a special condition: the angle of minimum deviation $\delta_m$ is exactly equal to the prism angle $A$.

At the position of minimum deviation, a light ray passes symmetrically through the prism. This symmetry dictates that the angle of incidence $i_1$ equals the angle of emergence $i_2$, and the two internal refracting angles are equal, meaning $r_1 = r_2 = \frac{A}{2}$.

The general formula for deviation is $\delta = i_1 + i_2 - A$. At minimum deviation, this simplifies to:

$$\delta_m = 2i_1 - A$$

Substituting our given condition $\delta_m = A$ into this equation, we get:

$$A = 2i_1 - A \implies 2i_1 = 2A \implies i_1 = A$$

Since $r_1 = \frac{A}{2}$, we can clearly see that $r_1 = \frac{i_1}{2}$. This elegantly proves that option (b) is absolutely correct!

### Unveiling the Refractive Index

Now, let's determine the refractive index $\mu$ of the prism material. We apply Snell's Law at the first refracting surface:

$$\sin i_1 = \mu \sin r_1$$

Substituting $i_1 = A$ and $r_1 = \frac{A}{2}$, we get:

$$\sin A = \mu \sin \left(\frac{A}{2}ight)$$

Using the double-angle trigonometric identity $\sin A = 2 \sin \left(\frac{A}{2}ight) \cos \left(\frac{A}{2}ight)$, we can expand the left side:

$$2 \sin \left(\frac{A}{2}ight) \cos \left(\frac{A}{2}ight) = \mu \sin \left(\frac{A}{2}ight)$$

Canceling the sine terms yields a beautiful relation for the refractive index:

$$\mu = 2 \cos \left(\frac{A}{2}ight)$$

This result immediately shows that option (d), which claims $A = \frac{1}{2}\cos^{-1}\left(\frac{\mu}{2}ight)$, is incorrect.

### The Grazing Emergence Condition

Let's shift our focus to option (c), which explores a completely different scenario: the emergent ray is tangential to the second surface. This is known as **grazing emergence**, where the angle of emergence $i_2 = 90^\circ$.

For grazing emergence, the angle of refraction at the second surface $r_2$ must be exactly equal to the critical angle $	heta_c$. Because the sum of the internal angles always equals the prism angle ($r_1 + r_2 = A$), our new $r_1$ becomes:

$$r_1 = A - 	heta_c$$

Let's apply Snell's Law at the first surface again for this new ray path:

$$\sin i_1 = \mu \sin r_1 = \mu \sin(A - 	heta_c)$$

Using the trigonometric identity for $\sin(A - B)$, we expand this to:

$$\sin i_1 = \mu (\sin A \cos 	heta_c - \cos A \sin 	heta_c)$$

### The Algebraic Climax

We know from the definition of the critical angle that $\sin 	heta_c = \frac{1}{\mu}$. Using the Pythagorean identity, we can find $\cos 	heta_c$:

$$\cos 	heta_c = \sqrt{1 - \sin^2 	heta_c} = \sqrt{1 - \frac{1}{\mu^2}} = \frac{\sqrt{\mu^2 - 1}}{\mu}$$

Substituting these into our expanded Snell's Law equation:

$$\sin i_1 = \mu \left( \sin A \frac{\sqrt{\mu^2 - 1}}{\mu} - \cos A \frac{1}{\mu} ight)$$

The $\mu$ terms elegantly cancel out, leaving us with:

$$\sin i_1 = \sin A \sqrt{\mu^2 - 1} - \cos A$$

Finally, we substitute our previously derived expression for the refractive index, $\mu = 2 \cos \left(\frac{A}{2}ight)$. This means $\mu^2 = 4 \cos^2 \left(\frac{A}{2}ight)$:

$$\sin i_1 = \sin A \sqrt{4 \cos^2 \left(\frac{A}{2}ight) - 1} - \cos A$$

Taking the inverse sine gives us the exact expression presented in option (c):

$$i_1 = \sin^{-1}\left[\sin A \sqrt{4\cos^2 \frac{A}{2} - 1} - \cos Aight]$$

Thus, option (c) is perfectly correct!

### A Note on the Isosceles Assumption

What about option (a)? We found that at minimum deviation, $i_1 = A$. For an isosceles prism, if the two base angles are equal (i.e., $\angle B = \angle C$), the ray inside at minimum deviation is indeed parallel to the base. While a prism can be isosceles in different ways, it is a standard convention in such physics problems to assume the symmetric case unless stated otherwise. Under this standard assumption, option (a) is also considered correct.

Animated Solution for Physics - Optics: For an isosceles prism of angle $A$ and refractive index $μ$ , it is found that the angle of minimum deviation $δ_{m} = A$ . Which of the following options is/are correct?

Select Answer:

Visualized Solution

Visualizing the Prism

Minimum Deviation Conditions

Finding $i_{1}$ and $r_{1}$

Refractive Index $μ$

Grazing Emergence

Angles for Grazing Emergence

Snell's Law for Grazing Ray

Substituting Critical Angle

Final Expression for $i_{1}$

The Isosceles Assumption

The Sigma Insight: Refraction and Dispersion through Prism

The Symmetry of Minimum Deviation

Unveiling the Refractive Index

The Grazing Emergence Condition

The Algebraic Climax

A Note on the Isosceles Assumption

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