A glass beaker has a solid, plano-convex base of refractive index 1.60, as shown in the figure. The

Q: Why did we assume the rays hit the mirror at 90 degrees?

For an image to form exactly on the object, the light rays must retrace their path. According to the laws of reflection, a ray retraces its path only if it strikes the mirror normally (perpendicularly), meaning the angle of incidence is zero.

Q: Why can't we just use the equivalent mirror formula $P = 2P_L + P_M$?

You absolutely can! The equivalent power of the system is $P_{eq} = 2P_{lens} + P_{mirror}$. Since the mirror is plane, $P_{mirror} = 0$, so $P_{eq} = 2P_{lens}$. The focal length of the equivalent mirror is $F_m = -1/P_{eq} = -F_{lens}/2$. For a mirror, the object must be at the center of curvature to form an image on itself, so $h = R = 2|F_m| = F_{lens}$. Both methods give the exact same result!

Q: How do we determine the signs of the radii of curvature?

We use the standard Cartesian sign convention. Assuming light travels downwards, the pole is at the surface. For the convex surface of the glass (bulging upwards), the center of curvature lies below it, in the direction of incident light. Hence, $R = +9$ cm.

Q: Is the liquid lens plano-convex or plano-concave?

The liquid fills the space above the glass. Its top surface is flat ($R = \infty$), and its bottom surface rests on the convex glass, making it concave from the perspective of the liquid. Therefore, it acts as a plano-concave lens.

Q: What if the bottom surface wasn't silvered?

If the bottom wasn't a mirror, the light would simply refract out of the beaker. The system would just be a combination of two lenses, and it would form a real or virtual image somewhere else, but it wouldn't reflect back to form an image onto itself.

Question

A glass beaker has a solid, plano-convex base of refractive index 1.60, as shown in the figure. The radius of curvature of the convex surface (SPU) is 9 cm, while the planar surface (STU) acts as a mirror. This beaker is filled with a liquid of refractive index n up to the level QPR. If the image of a point object O at a height of h (OT in the figure) is formed onto itself, then, which of the following option(s) is(are) correct?

Options:

(A) For n = 1.42, h = 50 cm.

(B) For n = 1.35, h = 36 cm.

(C) For n = 1.45, h = 65 cm.

(D) For n = 1.48, h = 85 cm.

Sigma Percentile · Accepted Answer

Welcome to a fascinating journey into the world of optics! This problem from JEE Advanced is a beautiful symphony of refraction, reflection, and geometric intuition. At first glance, it might look like a complex mess of multiple media, but once we decode the physical meaning behind the words, it unravels elegantly.

### The Magic of Retracing Rays

The entire problem hinges on one critical phrase: **"the image is formed onto itself."**

What does this physically mean? Imagine you are the object $O$, sending out rays of light. For those rays to come back and form an image exactly where you are standing, they must retrace their exact path.

Now, look at the bottom of our beaker. It's a flat, planar mirror. According to the laws of reflection, a light ray will only retrace its path if it strikes the mirror perfectly perpendicularly, meaning the angle of incidence is exactly $0^\circ$. If it hits at any other angle, it will bounce off in a different direction and the image will form somewhere else.

### Decoding the Optical System

So, we have established that the rays must travel straight down, parallel to the principal axis, when they hit the mirror.

But wait, the rays started from a point object $O$, meaning they were diverging! How did diverging rays become parallel? This is the magic of our lens system. The liquid and the glass together act as a combined optical lens.

For a lens system to convert diverging rays from a point into parallel rays, that point must be exactly at its **principal focus**. This is a massive conceptual breakthrough! It means the height $h$ of the object is simply the equivalent focal length $F_{eq}$ of the combined lens system.

$$h = F_{eq}$$

### The Lens Maker's Magic

Now, let's break down our optical system into two separate lenses and use the Lens Maker's Formula:

$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}ight)$$

First, the glass base. It is a plano-convex lens. The light hits the convex surface first. Using our sign convention (downward is positive), the center of curvature is below the surface, so $R_1 = +9 	ext{ cm}$. The bottom is flat, so $R_2 = \infty$.

$$\frac{1}{f_g} = (1.6 - 1)\left(\frac{1}{9} - \frac{1}{\infty}ight) = \frac{0.6}{9} = \frac{1}{15} 	ext{ cm}^{-1}$$

Next, the liquid lens. It fills the space above the glass. Its top surface is flat ($R_1 = \infty$), and its bottom surface rests on the convex glass, making it concave from the liquid's perspective ($R_2 = +9 	ext{ cm}$).

$$\frac{1}{f_l} = (n - 1)\left(\frac{1}{\infty} - \frac{1}{9}ight) = -\frac{n - 1}{9}$$

### Bringing It All Together

Since these two lenses are in contact, their equivalent power is simply the sum of their individual powers.

$$\frac{1}{F_{eq}} = \frac{1}{f_g} + \frac{1}{f_l}$$

Substituting our values:

$$\frac{1}{F_{eq}} = \frac{1}{15} - \frac{n - 1}{9}$$

To solve this, we take a common denominator of $45$:

$$\frac{1}{F_{eq}} = \frac{3 - 5(n - 1)}{45} = \frac{8 - 5n}{45}$$

Since we established earlier that $h = F_{eq}$, we get our master equation:

$$h = \frac{45}{8 - 5n}$$

### The Final Verification

The hard work is done! Now we just need to test the given options by plugging in the values of $n$.

For option (A), let $n = 1.42$:
$$h = \frac{45}{8 - 5(1.42)} = \frac{45}{8 - 7.1} = \frac{45}{0.9} = 50 	ext{ cm}$$
This matches perfectly!

For option (B), let $n = 1.35$:
$$h = \frac{45}{8 - 5(1.35)} = \frac{45}{8 - 6.75} = \frac{45}{1.25} = 36 	ext{ cm}$$
This also matches perfectly!

Testing options (C) and (D) with the same formula will yield incorrect heights. Thus, the correct options are indeed (A) and (B).

By trusting our physical intuition over blind formula application, we turned a complex multi-media problem into an elegant exercise in focal lengths. Always look for the physical story the problem is trying to tell you!

Select Answer:

Visualized Solution

$h$ and Image Formation

Condition for Retracing Path

Equivalent Lens Focus

Focal Length of Glass Lens

Focal Length of Liquid Lens

Combined Focal Length

Final Relation for $h$

Checking the Options

The Sigma Insight: Refraction at Spherical Surface

The Magic of Retracing Rays

Decoding the Optical System

The Lens Maker's Magic

Bringing It All Together

The Final Verification

Similar Questions

A point object is placed at the centre of a glass sphere of radius $6 cm$ and refractive index $1.5$ . The distance of the virtual image from the surface of the sphere is