Animated Solution for Physics - Optics: A transparent solid cylinder rod has a refractive index of $\frac{2}{\sqrt{3}}$. It is surrounded by air. A light ray is incident at the mid-point of one end of the rod as shown in the figure. The incident angle $\theta$ for which the light ray grazes along the wall of the rod is

Imagine a beam of light entering a transparent cylindrical rod. As it crosses the boundary from air into the denser medium of the rod, it bends—this is refraction in action. The problem states a fascinating condition: after this initial refraction, the light ray hits the top curved wall of the cylinder and grazes along it.

What does "grazing" mean in optics? It is the hallmark of the critical angle. When a light ray travels from a denser medium to a rarer medium and refracts exactly at $90^{\circ }$ along the boundary, the angle of incidence is called the critical angle, denoted by $C$.

To find this critical angle, we use the fundamental relationship between the critical angle and the refractive index $\mu$:

We are given that the refractive index of the rod is $\mu =\frac{2}{\sqrt{3}}$. Substituting this into our formula gives:

From our knowledge of trigonometry, we know that the angle whose sine is $\frac{\sqrt{3}}{2}$ is $60^{\circ }$. Therefore, the light ray strikes the top wall at exactly:

Now, let's look at the geometry inside the cylinder. The normal to the flat circular end is perfectly horizontal (along the axis of the cylinder), while the normal to the top curved wall is perfectly vertical.

These two normals intersect at a right angle ($90^{\circ }$). If we draw a right-angled triangle using the refracted ray, the horizontal axis, and the vertical normal, we can easily relate the angle of refraction $r$ at the first surface to the critical angle $C$ at the second surface.

Because the sum of the acute angles in a right-angled triangle is $90^{\circ }$, we have:

Since we already found that $C=60^{\circ }$, we can solve for $r$:

This tells us that the light ray refracted at an angle of $30^{\circ }$ when it first entered the cylinder. Let's also note that $\sin 30^{\circ }=\frac{1}{2}$, which will be very useful in the next step.

We are finally ready to find the incident angle $\theta$. We apply Snell's Law at the flat circular end where the light first enters from the air into the rod. Snell's Law states:

Here, the first medium is air (${\mu }_1=1$) and the angle of incidence is $\theta$. The second medium is the rod (${\mu }_2=\frac{2}{\sqrt{3}}$) and the angle of refraction is $r=30^{\circ }$. Substituting these values:

Now, plug in the value of $\sin 30^{\circ }$:

The $2$ in the numerator and the $2$ in the denominator cancel out beautifully, leaving us with:

To isolate $\theta$, we simply take the inverse sine:

This is the exact angle at which the light must enter the cylinder to eventually graze along its top wall. If the angle were any smaller, the ray would undergo total internal reflection—the very principle that makes optical fibers work!