Question 7

For the image to form on the object itself, light rays must strike the plane mirror (STU) normally. This means after refraction at the Air-Liquid and Liquid-Glass interfaces, the rays must be parallel to the principal axis.

Using the effective power of the silvered system or paraxial ray tracing:

1. Refraction at the flat liquid surface (at level QPR): The apparent position of O relative to the liquid surface is $h' = n \cdot h_{air}$.

2. Refraction at the curved interface (SPU) with radius $R = +9$ cm:

Using $\frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R}$.

For the ray to strike the mirror normally, it must be parallel to the axis after refraction, so $v = \infty$.

Here $\mu_2 = 1.60$, $\mu_1 = n$, and $u = -(h - \text{thickness})$.

Assuming the base is thin or $h$ is the distance to the effective surface:

$\frac{1.6}{\infty} - \frac{n}{-h} = \frac{1.6 - n}{9}$

$\frac{n}{h} = \frac{1.6 - n}{9} \Rightarrow h = \frac{9n}{1.6 - n}$ is not quite right as $h$ is measured from $T$.

Using the equivalent mirror approach where the object is at the center of curvature of the effective mirror:

$P_{eff} = 2P_{surf1} + 2P_{surf2} + P_{mirror}$.

$P_{surf1} (Air-Liq) = 0$ (flat).

$P_{surf2} (Liq-Glass) = \frac{1.6 - n}{9}$.

$P_{eff} = 2(0) + 2(\frac{1.6 - n}{9}) = \frac{3.2 - 2n}{9}$.

For image at object: $h = R_{eff} = \frac{2}{P_{eff}} = \frac{2}{(3.2 - 2n)/9} = \frac{9}{1.6 - n}$.

Testing the options:

(A) $n = 1.42 \Rightarrow h = \frac{9}{1.6 - 1.42} = \frac{9}{0.18} = 50$ cm. (Correct)

(B) $n = 1.35 \Rightarrow h = \frac{9}{1.6 - 1.35} = \frac{9}{0.25} = 36$ cm. (Correct)

(C) $n = 1.45 \Rightarrow h = \frac{9}{1.6 - 1.45} = \frac{9}{0.15} = 60$ cm. (Incorrect)

(D) $n = 1.48 \Rightarrow h = \frac{9}{1.6 - 1.48} = \frac{9}{0.12} = 75$ cm. (Incorrect)