Q.A magician during a show makes a glass lens with $n = 1.47$ disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water?

The key idea here is optical invisibility through index matching. When light passes from one medium to another, it bends (refracts) if the two media have different refractive indices. Some light is also reflected at the interface. If the refractive indices are exactly equal, there is no refraction and no reflection — the boundary becomes optically undetectable, and the object seems to vanish. This is why a glass rod can disappear in oil of the same refractive index.

Let’s work through the reasoning step by step.

1. Understand the condition for invisibility

   For a body to become invisible when placed in a liquid, light must pass through the interface between the body and the liquid without any deviation or reflection. This happens only when the refractive indices of the body and the liquid are identical. If they differ, some light is reflected (making the surface visible) and the rest is bent (distorting the image).

2. Identify the given refractive index of the body

   The problem states that the body has a refractive index $\mu = 1.4$. This is a fixed property of the material.

3. Apply the matching condition

   Since invisibility requires the liquid to have the same refractive index as the body, the liquid’s refractive index must also be $1.4$.

4. Check the options

   • (A) $\mu = 1$ — too low; the body would be visible. …

Lens-maker's equation: $\dfrac{1}{f}=\left(\dfrac{n_{lens}}{n_{medium}}-1\right)\left(\dfrac{1}{R_1}-\dfrac{1}{R_2}\right)$.

The lens behaves like a plane sheet when $f\to\infty$, i.e. the bracket $\left(\dfrac{n_{lens}}{n_{medium}}-1\right)=0$. …

1. The concept — critical angle.

For light going from a denser medium into a rarer one, total internal reflection begins at the critical angle $\theta_c$, defined by refraction at $90^\circ$:

where $n$ is the refractive index of the pair of media.

2. Why this makes $\theta_c$ depend on colour.

$\sin\theta_c = 1/n$ means $\theta_c$ decreases as $n$ increases. So the colour with the least refractive index has the greatest critical angle.

3. Dispersion — which colour has the least $n$?

In a normally dispersive transparent medium, refractive index decreases with increasing wavelength (Cauchy's relation $n = A + B/\lambda^2$):

$n_{\text{violet}} > n_{\text{blue}} > n_{\text{green}} > n_{\text{red}}$ …

The key idea here is Malus’s law, which governs how the intensity of polarised light changes when it passes through a polaroid. When unpolarised light hits the first polaroid, half the intensity gets through — that’s a fixed result, independent of the angle. After that, each subsequent polaroid reduces the intensity by a factor of $\cos^2\theta$, where $\theta$ is the angle between the pass axis of the incoming polarised light and the pass axis of the polaroid.

So the problem is just applying this twice: once for $P_1 \to P_2$, and once for $P_2 \to P_3$. The angles are both 45°, and $\cos^2 45° = \frac12$.

1. First polaroid $P_1$ — unpolarised light of intensity $I_0 = 128\ \text{W m}^{-2}$ is incident. For unpolarised light, only half the intensity passes through any polaroid, regardless of orientation. So the intensity after $P_1$ is:

2. Second polaroid $P_2$ — the light emerging from $P_1$ is now polarised along the pass axis of $P_1$. The angle between $P_1$ and $P_2$ is $45°$. By Malus’s law, the intensity after $P_2$ is:

3. Third polaroid $P_3$ — the light emerging from $P_2$ is polarised along the pass axis of $P_2$. The angle between $P_2$ and $P_3$ is again $45°$. Applying Malus’s law once more: …

The key is to understand why the ionosphere reflects radio waves. It is not a hard, mirror-like surface. Instead, the ionosphere is a region of the upper atmosphere where ultraviolet radiation from the Sun knocks electrons free from atoms, creating a plasma. This plasma has a refractive index that decreases as the electron density increases (and as the wave frequency decreases). For a typical radio wave sent upward, the refractive index gradually falls with altitude.

As the wave travels into this region of lower refractive index, it bends — just like light bending when it goes from water into air. If the angle of incidence is shallow enough, the bending becomes so severe that the wave is turned around completely and sent back to the ground. This is the essence of total internal reflection, but happening gradually over a distance, not at a single sharp interface.

Let’s compare each option:

1. Reflection by a plane mirror — This is a surface phenomenon. Light bounces off a smooth boundary because the electrons in the mirror’s surface re-radiate the wave. The ionosphere has no such sharp boundary; the change in refractive index is spread over many kilometres. So this is not the right analogy.

2. Total internal reflection in a mirage — In a desert mirage, the air near the ground is hotter and less dense, so its refractive index is lower than the cooler air above. Light from the sky travelling downward enters this hot layer at a shallow angle and bends upward — the path curves so much that the light appears to come from the ground, as if reflected. This is exactly the same physics as the ionosphere: a gradual decrease in refractive index with height causes a wave to bend back. This is the correct match. …