Question

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?

Answer 1

Hint To solve this question we need to use the lens maker’s formula. Applying the given condition on the focal length of the glass lens will give the value of the refractive index of the liquid.

Formula Used: The formula used to solve this question is given by
$\Rightarrow \dfrac{1}{f} = \left( {\dfrac{{{n_2}}}{{{n_1}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
Here $f$ is the focal length of a lens having refractive index of ${n_2}$ , placed in a medium of refractive index ${\mu _1}$ , and the radius of curvature of its surfaces are ${R_1}$ and ${R_2}$ .

Complete step by step answer
Let the refractive index of the liquid be ${n_1}$ .
An object is visible due to the light which passes through it and reaches the observer’s eyes. The light can pass through an object through two phenomena only; one is reflection, while the other is refraction. But if the light does not undergo any of these two phenomena when it interacts with an object, then the object will become invisible.
Now, we know that a glass lens refracts the light which is incident on it. This is due to the curvature of its surface. But if the surface becomes plane, then it will not refract the light. Instead the light will pass through it undisturbed. So the lens will become invisible.
From the lens maker’s formula we have
$\Rightarrow \dfrac{1}{f} = \left( {\dfrac{{{n_2}}}{{{n_1}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
If the glass lens is acting like a plane sheet, then its focal length tends to infinity. So the LHS of the above side, which is the reciprocal of the focal length, will become zero. So, we get
$\Rightarrow 0 = \left( {\dfrac{{{n_2}}}{{{n_1}}} - 1} \right)\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
The quantity $\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$ cannot be zero for any lens. So we have
$\Rightarrow \dfrac{{{n_2}}}{{{n_1}}} - 1 = 0$
$\Rightarrow \dfrac{{{n_2}}}{{{n_1}}} = 1$
So we get
$\Rightarrow {n_1} = {n_2}$
According to the question, the refractive index of the glass lens, ${n_2} = n = 1.47$ . So, we get the refractive index of the liquid as
$\Rightarrow {n_1} = 1.47$
Hence, the refractive index of the liquid is equal to $1.47$ .

Note
This question could also be attempted very easily by appreciating the fact that we are able to see two mediums distinctly because of the difference in their refractive indices. But if the two mediums having the same refractive index are placed together, then the combination becomes a homogenous, undistinguishable medium. So, we could directly jump to the conclusion that for making the glass lens invisible, it should be placed in a liquid having the same refractive index.