Question

A point light source $S$ is outside a cylinder on its axis near the end face (base). Determine the minimum refractive index $n$ of the cylinder material for which none of the rays entering the base will emerge from the lateral surface.

Answer 1

Hint: When light rays incident at an angle ${{90}^{o}}$, then only none of the rays will emerge from the lateral interface and then the angle of incidence $\alpha$ on the inner surface will satisfy the condition $\sin \alpha >\dfrac{1}{n}$.

Complete answer:
Generally refractive index is a numerical value calculated from the ratio of speed of light in vacuum to that in a second medium of higher density. This is because the speed of light varies for each medium. In this problem vacuum and cylindrical medium are considered.

If the connection holds for a ray with an angle of incidence $\alpha $of the angle on the inner surface $y \approx \dfrac{\pi }{2}$, In spite of the fact that materials' refractive indices are frequently quoted using a single value, none of the rays will exit the lateral surface of the cylinder, $\sin \alpha > \dfrac{1}{n}$.

In this instance, the lateral surface will experience an entire internal reflection of the ray. From geometrical reasoning, it follows that:
$\sin \alpha = \sqrt {1 - {{\sin }^2}\beta }$
$\sin \beta = \dfrac{1}{n}$
$\therefore {n_{\min }} = \sqrt 2$

Therefore, the minimum refractive index $n$ of the cylinder material is $\sqrt 2 $.

Note:Refractive index can also be calculated by Snell’s law, who explained the ratio of the sine of the angle of incidence and sine of the angle of refraction is identical to the ratio of the phase velocity in two different medium and also this is identical to ratio of indices of the refraction.