Animated Solution for Physics - Optics: A glass prism of refractive index 1.5 is immersed in water (refractive index 4/3). A light beam incident normally on the face $AB$ is totally reflected to reach the face $BC$ if

The phenomenon of Total Internal Reflection (TIR) is one of the most beautiful and practical concepts in optics. It is the underlying principle that allows fiber optic cables to transmit massive amounts of data across the globe at the speed of light, without losing any information along the way. In this problem, we are going to explore a classic scenario where a glass prism is submerged in water, and we need to determine the precise mathematical condition for a light ray to be totally internally reflected.

Imagine a glass prism with a refractive index of ${\mu }_g=1.5$. Instead of being surrounded by air, as is common in many textbook problems, this prism is completely immersed in water, which has a refractive index of ${\mu }_w=\frac{4}{3}$.

A light beam strikes the face $AB$ of the prism normally. What does it mean for a ray to strike "normally"? It means the ray is perfectly perpendicular to the surface, making an angle of incidence of $0^{\circ }$ with the normal line. According to Snell's Law, when light hits a boundary at a $90^{\circ }$ angle to the surface, it does not bend or refract. It marches straight through the interface and continues its journey inside the glass prism, completely undisturbed.

As the ray travels straight through the glass, it eventually hits the slanted face $AC$. To understand what happens at this boundary, we must first determine the angle of incidence.

Let's draw a normal—a perpendicular imaginary line—to the face $AC$ at the exact point where the ray strikes. The face $AB$ is perfectly vertical, and our incident ray is perfectly horizontal. The angle between the vertical face $AB$ and the slanted face $AC$ is given as $\theta$. By applying simple geometry and the properties of similar triangles, we can deduce that the angle between the horizontal incident ray and the normal to $AC$ is also exactly $\theta$.

This is a crucial step! In optics problems, getting the geometry right is half the battle won. If you miscalculate the angle of incidence, even the most flawless physics equations will lead you to the wrong answer.

Snell's Law and Critical Angle

Now, the light ray is inside the denser glass medium (${\mu }_g=1.5$) and is trying to escape into the rarer water medium (${\mu }_w=\frac{4}{3}$). However, light can be trapped inside the denser medium if it hits the boundary at a steep enough angle. This trapping mechanism is known as Total Internal Reflection (TIR).

For TIR to occur, the angle of incidence must be strictly greater than a specific threshold known as the critical angle, denoted by ${\theta }_c$. But what exactly is the critical angle? It is the specific angle of incidence for which the angle of refraction is exactly $90^{\circ }$, meaning the refracted ray grazes the boundary between the two media, unable to escape into the rarer medium.

We can find the critical angle using Snell's Law, which relates the angles of incidence and refraction to the refractive indices of the media:

Since the sine of $90^{\circ }$ is exactly $1$, the equation simplifies beautifully to:

This is our master equation for the problem! It tells us that the critical angle depends entirely on the ratio of the refractive indices of the two media involved.

With our master equation ready, all that remains is to plug in the numbers. We know the refractive index of water is ${\mu }_w=\frac{4}{3}$ and the refractive index of glass is ${\mu }_g=1.5$, which can be written as the fraction $\frac{3}{2}$.

Substituting these values into our equation, we get:

To divide these fractions, we multiply the numerator by the reciprocal of the denominator:

So, the sine of the critical angle is $\frac{8}{9}$.

For the ray to undergo Total Internal Reflection and reflect downwards to reach face $BC$, our angle of incidence $\theta$ must be strictly greater than the critical angle ${\theta }_c$.

Since the sine function is strictly increasing in the first quadrant (from $0^{\circ }$ to $90^{\circ }$), a larger angle will always yield a larger sine value. Therefore, we can take the sine of both sides without changing the inequality sign:

Substituting the value we just calculated, we arrive at our final condition:

And there we have it! By carefully combining the geometry of the prism with the physical principles of Snell's Law and Total Internal Reflection, we have successfully derived the exact condition required for the light ray to reach face $BC$. This elegant interplay between math and physics is what makes optics such a rewarding subject to study.