A ray of light travelling in a transparent medium falls on a surface separating the medium from air

Q: Why does the ray have to travel from a denser to a rarer medium for TIR?

When light travels from a denser to a rarer medium, it bends away from the normal. As the angle of incidence increases, the angle of refraction eventually reaches 90 degrees. If the light traveled from rarer to denser, it would bend towards the normal, and the angle of refraction could never reach 90 degrees.

Q: What happens if the angle of incidence is exactly equal to the critical angle?

If the angle of incidence is exactly equal to the critical angle, the refracted ray grazes the surface of the interface (angle of refraction is exactly 90 degrees). TIR strictly requires the angle of incidence to be greater than the critical angle.

Q: Why do we use the sine function to compare the angles?

We use the sine function because Snell's Law naturally relates the sine of the angles to the refractive indices. Since sine is an increasing function between 0 and 90 degrees, $i > \theta_c$ implies $\sin i > \sin \theta_c$.

Q: Can TIR happen if the second medium is not air?

Yes, absolutely! TIR can happen at any interface as long as the light is traveling from a medium with a higher refractive index to a medium with a lower refractive index. The critical angle formula would just become $\sin \theta_c = n_{rarer} / n_{denser}$.

Q: Why are there multiple correct options for this question?

The condition we derived is an inequality ($n > 1.414$), not a strict equation. Any refractive index value that is greater than 1.414 will cause TIR at a 45-degree angle of incidence. Both 1.5 and 1.6 satisfy this mathematical condition.

Question

A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence $45^\circ$. The ray undergoes total internal reflection. If $n$ is the refractive index of the medium with respect to air, select the possible value (s) of $n$ from the following

Options:

(A) 1.3

(B) 1.4

(C) 1.5

(D) 1.6

Sigma Percentile · Accepted Answer

### The Magic of Total Internal Reflection

Imagine you are swimming underwater in a crystal-clear pool at night. If you look straight up, you can see the stars. But if you look at the water's surface from a shallow angle, it suddenly turns into a perfect, silvery mirror, reflecting the bottom of the pool. This beautiful phenomenon is called **Total Internal Reflection (TIR)**, and it is the exact principle we are exploring in this problem.

### Analyzing the Setup

We are given a ray of light traveling inside a transparent medium, heading towards the boundary with air. The ray strikes this boundary at an angle of incidence $i = 45^\circ$.

The problem explicitly states that the ray undergoes Total Internal Reflection. This is our biggest clue. For TIR to happen, the light must be traveling from a denser medium to a rarer medium (which it is, from the medium to air), and the angle of incidence must be strictly greater than the **critical angle** $	heta_c$ of that interface.

### The Master Equation

Let's translate the physical condition into a mathematical inequality:

$$i > 	heta_c$$

Since the sine function is strictly increasing in the first quadrant (from $0^\circ$ to $90^\circ$), we can safely take the sine of both sides without changing the direction of the inequality:

$$\sin i > \sin 	heta_c$$

Now, what is $\sin 	heta_c$? According to Snell's Law, the sine of the critical angle is the ratio of the refractive index of the rarer medium to the denser medium. Since the rarer medium is air (with a refractive index of $1$), we have:

$$\sin 	heta_c = \frac{1}{n}$$

### Final Calculation

Let's substitute our known values into the inequality. We know the angle of incidence $i = 45^\circ$:

$$\sin 45^\circ > \frac{1}{n}$$

We know the exact trigonometric value for $\sin 45^\circ$ is $\frac{1}{\sqrt{2}}$. Plugging this in gives us:

$$\frac{1}{\sqrt{2}} > \frac{1}{n}$$

Here is where you must be careful not to make a silly algebraic mistake. When we take the reciprocal of both sides of an inequality involving positive numbers, the inequality sign flips. Therefore:

$$n > \sqrt{2}$$

### Conclusion

We know that the square root of $2$ is an irrational number approximately equal to $1.414$. So, our final condition for the refractive index is:

$$n > 1.414$$

Looking at the given options:
(a) $1.3$ is less than $1.414$ (Incorrect)
(b) $1.4$ is less than $1.414$ (Incorrect)
(c) $1.5$ is greater than $1.414$ (Correct)
(d) $1.6$ is greater than $1.414$ (Correct)

Thus, the possible values for the refractive index of the medium are **1.5** and **1.6**.

Select Answer:

Visualized Solution

Visualizing the Setup

Condition for Total Internal Reflection

Relating Critical Angle and Refractive Index

Setting Up the Inequality

Substituting the Values

Evaluating the Trigonometric Function

Solving for Refractive Index

Selecting the Correct Options

The Way Forward

The Sigma Insight: Refraction and Total Internal Reflection

The Magic of Total Internal Reflection

Analyzing the Setup

The Master Equation

Final Calculation

Conclusion

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