A thin convex lens L (refractive index = 1.5) is placed on a plane mirror M. When a pin is placed at

Q: Why must the rays strike the mirror normally?

For an image to coincide with the object, the light rays must retrace their exact path back to the source. A plane mirror only reflects rays back along their original path if they strike it at an angle of incidence of zero degrees, which means striking it normally (perpendicularly).

Q: Why is the liquid lens considered plano-concave?

The liquid fills the gap between the convex lens and the flat mirror. The top surface of the liquid takes the shape of the bottom of the convex lens (which is convex, making the liquid's top surface concave), and the bottom surface of the liquid rests on the flat mirror (making it plane). Hence, it's a plano-concave lens.

Q: Why is $R_1$ negative for the liquid lens?

According to the standard Cartesian sign convention, light travels from left to right (or top to bottom in this vertical setup). The center of curvature for the first concave surface of the liquid lens lies against the direction of incident light, making its radius negative.

Q: Can we assume the convex lens is equiconvex?

Yes, in standard JEE problems, unless the radii of curvature are specified to be different, a 'convex lens' is assumed to be an equiconvex lens where both surfaces have the same magnitude of radius $R$.

Q: What is the focal length of a plane mirror, and why don't we add its power?

The focal length of a plane mirror is infinity, meaning its optical power is zero. It doesn't converge or diverge the rays; it merely folds the optical path back on itself. That's why we only calculate the equivalent power of the lenses.

Question

A thin convex lens $L$ (refractive index = 1.5) is placed on a plane mirror $M$. When a pin is placed at $A$, such that $OA = 18$ cm, its real inverted image is formed at $A$ itself, as shown in figure. When a liquid of refractive index $\mu_l$ is put between the lens and the mirror, the pin has to be moved to $A'$, such that $OA' = 27$ cm, to get its inverted real image at $A'$ itself. The value of $\mu_l$ will be

Options:

(A) $\sqrt{3}$

(B) $\sqrt{2}$

(C) $\frac{4}{3}$

(D) $\frac{3}{2}$

Sigma Percentile · Accepted Answer

### The Magic of Auto-Collimation

Look closely at the setup. When the pin is placed at point $A$, its image is formed exactly back at $A$. This magical coincidence is known as auto-collimation. It is only possible if the light rays strike the plane mirror normally, meaning they become perfectly parallel after passing through the lens.

And when do rays become parallel? Exactly, when the object is placed at the principal focus! So, the focal length of our convex lens is simply the distance $OA$, which gives us:

$$f_1 = 18 	ext{ cm}$$

### The Hidden Liquid Lens

Now, we introduce a liquid between the lens and the mirror. This trapped liquid takes the shape of the gap, forming a new plano-concave lens. To get the image back at the object's position, the pin has to be moved to $A'$, which is $27 	ext{ cm}$ away.

This means the new focal length of the entire lens combination is $27 	ext{ cm}$. We know the formula for the equivalent focal length of lenses in contact:

$$\frac{1}{F_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$$

Let's substitute our known values into this raw structure:

$$\frac{1}{27} = \frac{1}{18} + \frac{1}{f_2}$$

From here, we can easily calculate the focal length of the liquid lens. Taking the LCM and solving, we get:

$$\frac{1}{f_2} = \frac{1}{27} - \frac{1}{18} = \frac{2 - 3}{54} = -\frac{1}{54}$$

$$f_2 = -54 	ext{ cm}$$

The negative sign perfectly confirms it's a diverging, plano-concave lens.

### Unlocking the Radius

Now, recall the famous Lens Maker's formula. For our equiconvex glass lens, we will use this to find the radius of curvature of the lens surfaces:

$$\frac{1}{f_1} = (\mu_g - 1)\left(\frac{2}{R}ight)$$

Substituting the values for the glass lens:

$$\frac{1}{18} = (1.5 - 1)\left(\frac{2}{R}ight)$$

$$\frac{1}{18} = 0.5 	imes \frac{2}{R} \Rightarrow R = 18 	ext{ cm}$$

### The Final Refractive Index

Don't make a silly mistake here. The liquid lens is plano-concave. Its first surface is concave, taking a radius of $-R$, and the second surface is perfectly flat, taking a radius of $\infty$. Let's carefully substitute these into the formula for the liquid lens:

$$\frac{1}{f_2} = (\mu_l - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}ight)$$

$$-\frac{1}{54} = (\mu_l - 1)\left(\frac{1}{-18} - \frac{1}{\infty}ight)$$

The minus signs on both sides will elegantly cancel out:

$$\mu_l - 1 = \frac{18}{54} = \frac{1}{3}$$

Moving the one over, we get our final answer:

$$\mu_l = 1 + \frac{1}{3} = \frac{4}{3}$$

This is a highly classic JEE problem that beautifully merges the concepts of auto-collimation and lens combinations. The refractive index of the liquid is exactly $4/3$, which happens to be the refractive index of water.

Select Answer:

Visualized Solution

$f_{1}$ of Convex Lens

$F_{e q}$ of Combination

Combination Formula

Calculating $f_{2}$

Radius of Convex Lens

Lens Maker for Liquid Lens

Solving for $μ_{l}$

Final Answer

The Sigma Insight: Refraction at Spherical Surface

The Magic of Auto-Collimation

The Hidden Liquid Lens

Unlocking the Radius

The Final Refractive Index

Similar Questions

In the figure, light is incident on a thin lens as shown. The radius of curvature for both the surfaces is $R$ . Determine the focal length of this system.

A quarter cylinder of radius $R$ and refractive index $1.5$ is placed on a table. A point object $P$ is kept at a distance of $m R$ from it. Find the value of $m$ for which a ray from $P$ will emerge parallel to the table as shown in figure.