Animated Solution for Physics - Optics: A point source $S$ is placed at the bottom of a transparent block of height $10\text{ mm}$ and refractive index $2.72$. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter $11.54\text{ mm}$ on the top of the block. The refractive index of the liquid is

Have you ever looked up from underwater and noticed that the entire world above the surface is compressed into a circular window of light? This fascinating optical phenomenon is known as Snell's Window. In this problem, we are exploring a miniature version of this effect. We have a point source of light placed at the bottom of a transparent block, and we are observing the circular bright spot it forms on the top surface.

Imagine the light rays emanating from the point source $S$ at the bottom of the block. As these rays travel upwards and hit the boundary between the block and the surrounding liquid, they undergo refraction. Because the block has a higher refractive index than the liquid, the rays bend away from the normal.

As the angle of incidence increases, the angle of refraction also increases. Eventually, we reach a specific angle of incidence where the refracted ray grazes the surface of the block, meaning the angle of refraction is exactly $90^{\circ }$. This specific angle is called the critical angle, denoted by ${\theta }_C$. Any ray hitting the surface at an angle greater than ${\theta }_C$ will undergo total internal reflection and bounce back into the block. Therefore, the light that successfully escapes into the liquid forms a bright circular spot, and the edge of this spot corresponds exactly to the rays hitting the surface at the critical angle.

To find the relationship between the critical angle and the refractive indices of the two media, we turn to Snell's Law. At the edge of the bright spot, the angle of incidence is ${\theta }_C$ and the angle of refraction is $90^{\circ }$.

Applying Snell's Law at this boundary:

Since $\sin 90^{\circ }=1$, we can simplify this to:

This elegant equation tells us that the sine of the critical angle is simply the ratio of the refractive index of the rarer medium (the liquid) to the denser medium (the block).

Now, let's connect this optical principle to the physical dimensions given in the problem. If we draw a cross-section of the setup, we can form a right-angled triangle $\Delta PQS$. The vertices are the point source $S$, the center of the bright spot $P$ directly above it, and a point $Q$ on the edge of the bright spot.

The height of the block is $PS=h$, and the radius of the bright spot is $PQ=r$. The hypotenuse of this triangle is the path of the light ray, $SQ$. Using the Pythagorean theorem, the length of the hypotenuse is:

By alternate interior angles, the angle $\angle PSQ$ is equal to the critical angle ${\theta }_C$. Looking at our right-angled triangle, we can express the sine of this angle as the ratio of the opposite side to the hypotenuse:

We now have two different expressions for $\sin {\theta }_C$. By equating them, we bridge the gap between the optical properties and the physical geometry:

Our goal is to find the refractive index of the liquid, ${\mu }_l$. Rearranging the equation, we get:

Let's plug in the numbers provided in the problem. The diameter of the bright spot is $11.54\text{ mm}$, so the radius $r$ is half of that:

The height of the block is $h=10\text{ mm}$, and the refractive index of the block is ${\mu }_{\text{block}}=2.72$. Substituting these values into our master equation:

Let's calculate the denominator first:

Now, substituting this back:

Notice how beautifully the numbers align! The ratio $\frac{5.77}{11.545}$ is exactly $0.5$.

And there we have it! The refractive index of the liquid is 1.36. This problem is a fantastic demonstration of how abstract optical laws manifest in measurable, physical phenomena.