Question

Light traveling through three transparent substances follows the path shown in the figure. Assuming that total internal reflection does take place on the bottom surface of medium 2, arrange the refractive index in the increasing order.

• A. $\mu_1 < \mu_2 < \mu_3$
• B. $\mu_2 < \mu_1 < \mu_3$
• C. $\mu_1 < \mu_3 < \mu_2$
• D. $\mu_3 < \mu_1 < \mu_2$

Answer 1

Answer: (D)

(D)

Answer 2

To determine the increasing order of refractive indices $\mu_1, \mu_2, \mu_3$, we analyze the path of light at each interface using Snell's Law and the conditions for Total Internal Reflection (TIR).

Let $\theta_1$ be the angle of incidence in medium 1 at the interface with medium 2.
Let $\theta_2$ be the angle of refraction in medium 2 at the interface with medium 1. Due to parallel layers, $\theta_2$ is also the angle of incidence in medium 2 at the interface with medium 3.

1. Refraction at the interface between Medium 1 and Medium 2:

From the figure, the light ray bends towards the normal as it passes from medium 1 to medium 2.
When light bends towards the normal, it implies that the light is moving from a rarer medium to a denser medium.
Therefore, the refractive index of medium 2 is greater than that of medium 1.
This gives us the first inequality: $\mu_2 > \mu_1$

According to Snell's Law at this interface: $\mu_1 \sin \theta_1 = \mu_2 \sin \theta_2$ Since the ray bends towards the normal, $\theta_1 > \theta_2$.
If $\theta_1 > \theta_2$, then $\sin \theta_1 > \sin \theta_2$.
For the equality $\mu_1 \sin \theta_1 = \mu_2 \sin \theta_2$ to hold, it must be that $\mu_2 > \mu_1$. This confirms our initial observation.

2. Total Internal Reflection (TIR) at the interface between Medium 2 and Medium 3:

The problem statement explicitly mentions that total internal reflection takes place on the bottom surface of medium 2 (which is the interface between medium 2 and medium 3).
For TIR to occur, two conditions must be met: (a) Light must travel from a denser medium to a rarer medium.
(b) The angle of incidence must be greater than the critical angle.

From condition (a), for TIR to occur when light travels from medium 2 to medium 3, medium 2 must be denser than medium 3.
Therefore, the refractive index of medium 2 is greater than that of medium 3.
This gives us the second inequality: $\mu_2 > \mu_3$

Now we have established: $\mu_2 > \mu_1$
$\mu_2 > \mu_3$

This means $\mu_2$ is the largest refractive index among the three. We need to compare $\mu_1$ and $\mu_3$.

From condition (b) for TIR, the angle of incidence $\theta_2$ must be greater than the critical angle $C_{23}$ for the interface between medium 2 and medium 3.
The critical angle $C_{23}$ is given by $\sin C_{23} = \frac{\mu_3}{\mu_2}$.
So, $\theta_2 > C_{23}$, which implies $\sin \theta_2 > \sin C_{23}$.
$\sin \theta_2 > \frac{\mu_3}{\mu_2}$

From Snell's Law at the first interface, we have $\sin \theta_2 = \frac{\mu_1}{\mu_2} \sin \theta_1$.
Substitute this into the TIR condition: $\frac{\mu_1}{\mu_2} \sin \theta_1 > \frac{\mu_3}{\mu_2}$
Since $\mu_2$ is a positive refractive index, we can multiply both sides by $\mu_2$: $\mu_1 \sin \theta_1 > \mu_3$

Now, let's analyze this inequality to compare $\mu_1$ and $\mu_3$:
Since $\theta_1$ is an angle of incidence (and not 90 degrees for the ray to travel through the medium), $0 < \sin \theta_1 \le 1$.
If $\mu_1 = \mu_3$, then $\mu_1 \sin \theta_1 > \mu_1$. This implies $\sin \theta_1 > 1$, which is impossible. So $\mu_1 \ne \mu_3$.
If $\mu_1 < \mu_3$, then $\mu_1 \sin \theta_1 > \mu_3$ would imply $\sin \theta_1 > \frac{\mu_3}{\mu_1}$. Since $\mu_1 < \mu_3$, $\frac{\mu_3}{\mu_1} > 1$. So $\sin \theta_1 > (\text{a value greater than 1})$, which is impossible. So $\mu_1$ cannot be less than $\mu_3$.

The only remaining possibility is $\mu_1 > \mu_3$.

Combining all inequalities: $\mu_2 > \mu_1$
$\mu_1 > \mu_3$
$\mu_2 > \mu_3$

Therefore, the increasing order of refractive indices is $\mu_3 < \mu_1 < \mu_2$.