consider the refraction at the first surface of the prism where light enters. According to Snell's law:

$$n_i \sin(\theta_i) = n_r \sin(\theta_r)$$

where $ n_i $ is the refractive index of the medium the light is coming from (which we assume to be air, so $ n_i = 1 $), $ \theta_i $ is the angle of incidence $(60^{\circ})$, $ n_r $ is the refractive index of the prism, and $ \theta_r $ is the angle of refraction.

We have:

$$\sin(60^{\circ}) = n \sin(\theta_r)$$

$$\frac{\sqrt{3}}{2} = n \sin(\theta_r)$$

Next, we need to find the relationship between $ \theta_r $ and the internal angles of the prism. Since the prism angle is $30^{\circ}$ and the emergent ray is normal to the surface (meaning the angle of emergence is $0^{\circ}$), we can use the fact that the sum of angles inside the prism should be equal to $90^{\circ}$ (since the light travels through the prism and the final deviation angle should cancel out the prism angle). Therefore:

$$\theta_r + \phi = 30^{\circ}$$

Where $ \phi $ is the internal angle of deviation. Given that $ \phi + 0^{\circ} = 30^{\circ} $, it means $ \theta_r $ should equal $ 30^{\circ} $.

Therefore we have:

$$\sin(30^{\circ}) = \frac{1}{2}$$

Substituting $ \theta_r = 30^{\circ} $ into Snell's law equation:

$$\frac{\sqrt{3}}{2} = n \cdot \frac{1}{2}$$

Thus, solving for $n $, we find:

$$n = \sqrt{3}$$

Hence, the refractive index of the prism is:

Option A: $$\sqrt{3}$$

Ray Optics

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