Question

A prism of prism angle $(A=60^\circ)$ and refractive index $n_2$ is placed in a liquid of refractive index $n_1$. A light ray is incident on face $AB$ at constant angle of incidence $i$ and it emerges at surface $AC$ at angle of emergent $e$. Somehow $n_1$ is increasing at rate $N_1$ and $n_2$ is increasing at rate $N_2$. Choose the correct option(s)

• A. If $\frac{N_1}{N_2}>\frac{n_1}{n_2}$ then e is increasing
• B. If $\frac{N_1}{N_2}<\frac{n_1}{n_2}$ then e is increasing
• C. If $\frac{N_1}{N_2}>\frac{n_1}{n_2}$ then e is decreasing
• D. If $\frac{N_1}{N_2}=\frac{n_1}{n_2}$ then e is decreasing

Answer 1

Answer: (B), (C)

Options 2 and 3

Answer 2

We start with the two refraction conditions:

1. At the first face (AB): $n_1 \sin i = n_2 \sin r_1$

2. At the second face (AC): $n_2 \sin r_2 = n_1 \sin e$

Also, by geometry inside the prism: $r_1 + r_2 = A$ (with $A = 60^\circ$)

Since the angle of incidence $i$ is fixed, differentiate equation (1) with respect to time $t$: $n_1 \sin i = n_2 \sin r_1$

Differentiating (with $i$ constant) gives: $0 = n_2 \cos r_1 (dr_1/dt)$

More precisely, $\cos r_1 (dr_1/dt) = \sin i \cdot [ (d(n_1)/dt \cdot n_2 – n_1 \cdot d(n_2)/dt] / n_2^2$

Similarly, differentiating (2) and using $r_2 = A – r_1$, one eventually finds that $\cos e (de/dt) = (N_2n_1 – n_2N_1) \cdot [positive factors]$.

All the trigonometric factors (like $\sin(A − r_1)$, $\cos e$, $\cos r_1$, etc.) are positive in the allowed range. Thus the sign of $de/dt$ is determined by the factor $(N_2n_1 – n_2N_1)$. In other words,

$de/dt > 0$ if $N_2n_1 – n_2N_1 > 0 \implies N_1/N_2 < n_1/n_2$,

$de/dt < 0$ if $N_2n_1 – n_2N_1 < 0 \implies N_1/N_2 > n_1/n_2$.

Thus we conclude:

• If $(N_1/N_2) < (n_1/n_2)$ then the emergent angle $e$ is increasing.
• If $(N_1/N_2) > (n_1/n_2)$ then the emergent angle $e$ is decreasing.