Sigma Percentile
JEE Main 2003
LEVELJEE Main

Animated Solution for Physics - Optics: A ray of light is incident at the glass-water interface at an angle , it emerges finally parallel to the surface of water, then the value of would be

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Visualized Solution

The Sigma Insight: Refraction and Total Internal Reflection

Solution Diagram
The problem of a light ray passing through multiple parallel media is a classic in optics. It beautifully demonstrates how Snell's Law can be applied sequentially, and more importantly, how it can be simplified.

Tracing the Light Ray

Imagine you are following the journey of a single photon. It originates in the dense medium of glass, traveling upwards until it strikes the boundary with water. Here, it bends. It then travels through the water and hits the final boundary with air. The problem states a very specific condition for this final emergence: the ray travels parallel to the water surface. This means the angle of refraction in the air is exactly .

The Master Equation

Snell's Law
To understand what happens at each boundary, we rely on Snell's Law, which states that the product of the refractive index and the sine of the angle with the normal is constant across an interface:
Let's apply this to the first interface (glass to water):
Now, let's look at the second interface (water to air). Because the two boundaries are parallel, the normal lines are also parallel. By the geometric property of alternate interior angles, the angle of incidence at the second boundary is exactly . Applying Snell's Law here gives:

The Intermediate Layer Illusion

Notice something magical? Both expressions are equal to . This allows us to completely bypass the water layer and directly equate the initial state in the glass to the final state in the air:
This is a profound principle in optics: When light passes through a series of parallel transparent media, the intermediate layers do not affect the relationship between the initial angle of incidence and the final angle of emergence. The refractive index of water () was a distractor!

Final Calculation

We know that the refractive index of air () is , and is also . Substituting these values into our simplified equation:
Solving for the refractive index of the glass, we get:
And there we have it! A seemingly complex three-medium problem elegantly collapses into a single, simple equation.

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