Question 5: Comparing Refractive Indices
Step-by-Step Analysis
1. Interface 1-2 (Refraction):
The ray travels from medium 1 ($\mu_1$) to medium 2 ($\mu_2$). We can see the ray becomes steeper, meaning it bends towards the normal. This indicates that medium 2 is optically denser than medium 1.
$$ \mu_2 > \mu_1 $$
2. Interface 2-3 (Total Internal Reflection):
The ray hits the interface between medium 2 ($\mu_2$) and medium 3 ($\mu_3$) and reflects back into medium 2. Since no light enters medium 3, this is Total Internal Reflection (TIR). TIR can only happen when light travels from a denser medium to a rarer medium.
$$ \mu_2 > \mu_3 $$
From these two points, we know $\mu_2$ is the largest refractive index.
3. Comparing $\mu_1$ and $\mu_3$:
For TIR to occur at the second interface, the angle of incidence $i_2$ must be greater than the critical angle $C$.
$$ \sin(i_2) > \frac{\mu_3}{\mu_2} \implies \mu_2 \sin(i_2) > \mu_3 $$
Using Snell’s Law at the first interface, $\mu_1 \sin(i_1) = \mu_2 \sin(r_1)$.
From geometry (alternate interior angles), the angle of refraction at the first interface $r_1$ equals the angle of incidence at the second interface $i_2$.
$$ \mu_1 \sin(i_1) = \mu_2 \sin(i_2) $$
Substituting this into the TIR condition:
$$ \mu_1 \sin(i_1) > \mu_3 $$
Since $\sin(i_1) < 1$, it follows that $\mu_1 > \mu_1 \sin(i_1) > \mu_3$.
Therefore:
$$ \mu_1 > \mu_3 $$
Correct Option: (a) $\mu_3 < \mu_1 < \mu_2$