A thin biconvex lens of refractive index 3/2 is placed on a horizontal plane mirror as shown in the

Q: Why do the rays need to strike the mirror normally?

For the image to form exactly at the object's position, the light rays must retrace their entire path backwards. A plane mirror only reflects rays back along their original path if they hit it at a 90-degree angle (normally).

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

The liquid is trapped between the flat plane mirror and the curved bottom surface of the biconvex lens. Therefore, its bottom surface is plane ($R = \infty$) and its top surface takes the shape of the lens, making it concave upwards.

Q: Why do we add the powers of the lenses?

When thin lenses are placed in direct contact, their optical powers add up algebraically. This is a fundamental principle of combined optical systems, allowing us to treat them as a single equivalent lens.

Q: What is the sign convention for the radius of curvature?

Following the Cartesian sign convention, light travels from top to bottom. The top surface of the biconvex lens is convex upwards (positive $R$), and the bottom surface is convex downwards (negative $R$). For the liquid lens, the top surface is concave upwards (negative $R$).

Q: Does the distance between the lens and mirror matter?

In this specific problem, the lens is placed *on* the mirror, so the distance is effectively zero. The liquid fills this infinitesimally thin gap, forming a thin lens itself. If there were a significant gap, we would have to account for the separation distance in our equivalent focal length formula.

Question

A thin biconvex lens of refractive index $3/2$ is placed on a horizontal plane mirror as shown in the figure. The space between the lens and the mirror is then filled with water of refractive index $4/3$. It is found that when a point object is placed $15	ext{ cm}$ above the lens on its principal axis, the object coincides with its own image. On repeating with another liquid, the object and the image again coincide at a distance $25	ext{ cm}$ from the lens. Calculate the refractive index of the liquid.

Sigma Percentile · Accepted Answer

The behavior of light when it encounters a combination of lenses and mirrors is one of the most fascinating topics in optics. In this problem, we are presented with a beautiful setup: a biconvex glass lens resting on a plane mirror, with a liquid trapped in between.

This trapped liquid naturally forms a plano-concave lens. Thus, our system is effectively a combination of two lenses placed directly on a mirror.

### Analyzing the Setup

The most crucial piece of information given is that the object coincides with its own image. What does this physically mean?

For an image to form exactly at the object's position, the light rays must retrace their entire path backwards. A plane mirror only reflects rays back along their original path if they strike it at a perfect $90^\circ$ angle (normally).

If the rays are hitting the mirror normally, it implies that they emerged from the lens system parallel to the principal axis. And we know from the fundamental principles of optics that rays emerge parallel only when the object is placed exactly at the **focal point** of the lens system!

Therefore, the given object distances ($15	ext{ cm}$ and $25	ext{ cm}$) are actually the **equivalent focal lengths** of the combined lens systems in the two different cases.

### The Master Equation

Let's break down the system into its individual components. We have a solid biconvex glass lens and a liquid plano-concave lens. When thin lenses are in contact, their optical powers add up algebraically:

$$ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} $$

Let the radius of curvature of both surfaces of the biconvex lens be $R$. Using the lens maker's formula for the glass lens ($\mu_1 = \frac{3}{2}$):

$$ \frac{1}{f_1} = \left(\frac{3}{2} - 1ight)\left(\frac{1}{R} - \frac{1}{-R}ight) = \frac{1}{R} $$

Now, let's look at the water lens ($\mu_2 = \frac{4}{3}$). Its top surface matches the glass lens ($R_1 = -R$), and its bottom surface is flat against the mirror ($R_2 = \infty$):

$$ \frac{1}{f_2} = \left(\frac{4}{3} - 1ight)\left(\frac{1}{-R} - \frac{1}{\infty}ight) = -\frac{1}{3R} $$

In the first case, the combined focal length $F$ is $15	ext{ cm}$. Adding the powers:

$$ \frac{1}{15} = \frac{1}{R} - \frac{1}{3R} = \frac{2}{3R} $$

Solving this beautifully simple equation gives us the intrinsic geometry of the glass lens:

$$ R = 10	ext{ cm} $$

### Final Calculation

Now, the water is replaced by a mysterious new liquid of refractive index $\mu$. The new liquid lens has a focal length $f_3$:

$$ \frac{1}{f_3} = (\mu - 1)\left(\frac{1}{-R} - \frac{1}{\infty}ight) = -\frac{\mu - 1}{R} $$

With this new liquid, the object coincides with its image at $25	ext{ cm}$, meaning our new combined focal length $F'$ is $25	ext{ cm}$. Let's set up the equation again:

$$ \frac{1}{25} = \frac{1}{f_1} + \frac{1}{f_3} $$

$$ \frac{1}{25} = \frac{1}{R} - \frac{\mu - 1}{R} = \frac{2 - \mu}{R} $$

We are in the endgame now! We already know $R = 10	ext{ cm}$. Substituting this value into our equation:

$$ \frac{1}{25} = \frac{2 - \mu}{10} $$

Cross-multiplying to solve for $\mu$:

$$ 10 = 25(2 - \mu) $$

$$ 10 = 50 - 25\mu $$

$$ 25\mu = 40 $$

$$ \mu = \frac{40}{25} = 1.6 $$

And there we have it! By carefully analyzing the retracing of rays and systematically applying the lens maker's formula, we have successfully deduced the refractive index of the unknown liquid to be **$1.6$**.

Enter Numerical Value:

Visualized Solution

$Analyzing the Optical System$

$Focal Length of the Biconvex Lens$

$Focal Length of the Water Lens$

$Combined Focal Length (Case 1)$

$Calculating Radius of Curvature R$

$Focal Length of the New Liquid Lens$

$Combined Focal Length (Case 2)$

$Calculating the Unknown Refractive Index$

The Sigma Insight: Lens

Analyzing the Setup

The Master Equation

Final Calculation

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Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Optical System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Focal Length of the Biconvex Lens

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Focal Length of the Water Lens

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Combined Focal Length (Case 1)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Radius of Curvature R

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Focal Length of the New Liquid Lens

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Combined Focal Length (Case 2)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Unknown Refractive Index