Animated Solution for Physics - Optics: A light ray enters a solid glass sphere of refractive index $\mu =\sqrt{3}$ at an angle of incidence $60^{\circ }$. The ray is both reflected and refracted at the farther surface of the sphere. The angle (in degree) between the reflected and refracted rays at this surface is ……… .

A Journey Through a Glass Sphere

Imagine a beam of light traveling through the air, suddenly encountering a perfectly spherical drop of glass. What happens next is a beautiful dance of physics and geometry. In this problem, we are given a solid glass sphere with a refractive index of $\mu =\sqrt{3}$. A light ray strikes its surface at an angle of incidence of $60^{\circ }$. Our goal is to track this ray as it enters the sphere, travels to the other side, and splits into a reflected and a refracted ray. We need to find the exact angle between these two final rays.

Entering the Sphere

As the light ray hits the first surface, it moves from a rarer medium (air, $\mu =1$) into a denser medium (glass, $\mu =\sqrt{3}$). According to Snell's Law, the ray will bend towards the normal. Let's set up the equation to find the exact angle of refraction, $r_1$:

We know that $\sin (60^{\circ })=\frac{\sqrt{3}}{2}$. Substituting this into our equation gives:

The $\sqrt{3}$ terms cancel out beautifully on both sides, leaving us with:

This means our angle of refraction is exactly $r_1=30^{\circ }$. The ray now travels through the interior of the glass sphere at this angle relative to the normal.

Geometry of the Sphere

Here is where the elegant geometry of the sphere comes into play. The ray travels in a straight line from the first point of contact to the second point on the opposite side. If we draw lines from the center of the sphere to these two points, we form a triangle.

Because both of these lines are radii of the sphere, they are equal in length. This makes the triangle an isosceles triangle. In an isosceles triangle, the angles opposite the equal sides are also equal. Therefore, the angle of incidence at the second surface, $i_2$, must be exactly equal to the angle of refraction at the first surface:

Reflection and Refraction

At this second surface, the problem states that the ray undergoes both reflection and refraction. Let's analyze them one by one.

First, the reflection. According to the law of reflection, the angle of reflection must equal the angle of incidence. Since the ray is reflecting back inside the sphere, we measure this angle against the inward normal. The reflected ray bounces back at an angle of:

Next, the refraction. The ray is now trying to exit the glass and re-enter the air. We apply Snell's Law once more, this time moving from glass to air:

Substituting $\sin (30^{\circ })=\frac{1}{2}$, we get:

This tells us that the angle of emergence, $r_2$, is exactly $60^{\circ }$. As expected, the ray bends away from the normal as it enters the less dense air.

Finding the Angle

We now have all the pieces of the puzzle. We need to find the angle $\theta$ between the reflected ray (inside the sphere) and the refracted ray (outside the sphere).

Visualize the normal line at the second surface; it is a straight line representing $180^{\circ }$. The reflected ray is on one side of the boundary, making a $30^{\circ }$ angle with the normal. The refracted ray is on the other side of the boundary, making a $60^{\circ }$ angle with the normal.

The total angle between them is simply the straight line minus these two angles:

The reflected and refracted rays are perfectly perpendicular to each other! This is a stunning geometric result that emerges from the symmetric properties of the sphere and the specific refractive index provided.