One plano-convex and one plano-concave lens of same radius of curvature R but of different materials

Q: Why is the radius of the plane surface taken as infinity?

A plane surface can be thought of as a part of a sphere with an infinitely large radius. Mathematically, as $R \to \infty$, the curvature becomes zero, which perfectly describes a flat plane.

Q: Why do we take $R_2 = -R$ for the first lens?

According to the Cartesian sign convention, the direction of incident light is taken as positive (usually left to right). The center of curvature for the curved surface of the first lens lies to its left. Since we measure from the optical center against the direction of light, the radius is negative.

Q: Why is $R_1 = -R$ for the second lens?

The light enters the second lens through the curved surface. This is the exact same physical boundary as the second surface of the first lens. Its center of curvature is still to the left, against the incident light, so $R_1$ is $-R$.

Q: Can we just use a single formula for the whole system instead of splitting it?

Yes, you can use the general formula for a thick lens or a composite system, but for thin lenses in tight contact, adding their individual powers ($\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$) is the most straightforward and error-free method.

Q: What happens if we flip the entire combination around?

If you flip the combination, the light hits the plane surface of lens 2 first. The sign conventions will flip (e.g., centers of curvature will be on the right, so $+R$), but the final equivalent focal length will remain exactly the same. The optical power of a thin lens system is independent of which side the light enters from.

Question

One plano-convex and one plano-concave lens of same radius of curvature $R$ but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is $\mu_1$ and that of 2 is $\mu_2$, then the focal length of the combination is

Options:

(A) $\frac{2R}{\mu_1 - \mu_2}$

(B) $\frac{R}{2 - (\mu_1 - \mu_2)}$

(C) $\frac{R}{2(\mu_1 - \mu_2)}$

(D) $\frac{R}{\mu_1 - \mu_2}$

Sigma Percentile · Accepted Answer

### The Setup: Two Lenses in Harmony

Imagine you are an optical engineer handed two distinct pieces of glass. One is a plano-convex lens made of a material with refractive index $\mu_1$. The other is a plano-concave lens made of a different material with refractive index $\mu_2$. They are crafted so perfectly that their curved surfaces share the exact same radius of curvature, $R$. When you press them together, they form a seamless rectangular block with a curved boundary hidden inside.

Our mission is to find the equivalent focal length $f_{eq}$ of this combined system. To do this, we will treat the system as two separate thin lenses placed in tight contact and use the principle of superposition of optical powers.

### The Master Equation: Lens Maker's Formula

The fundamental tool we need is the **Lens Maker's Formula**, which relates the focal length $f$ of a lens to its refractive index $\mu$ and the radii of curvature of its two surfaces, $R_1$ and $R_2$:

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

We will also use the combination formula for thin lenses in contact, which states that their optical powers add up linearly:

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

### Analyzing the First Lens

Let's isolate the first lens (the plano-convex one on the left). Assume light enters from the left.

The first surface the light encounters is perfectly flat. A flat plane can be thought of as a sphere with an infinitely large radius, so **$R_1 = \infty$**.

The second surface is curved. If you trace the circle that forms this curve, its center lies to the left of the surface. Since incident light travels from left to right, we must measure the radius against the direction of light. By the Cartesian sign convention, this makes the radius negative. Thus, **$R_2 = -R$**.

Plugging these into the Lens Maker's formula:

$$\frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right)$$

Since $\frac{1}{\infty} = 0$, this simplifies beautifully to:

$$\frac{1}{f_1} = \frac{\mu_1 - 1}{R}$$

### Analyzing the Second Lens

Now, let's look at the second lens (the plano-concave one on the right).

The light first hits the curved surface. This is the exact same physical boundary we just analyzed, so its center of curvature is still to the left. Therefore, **$R_1 = -R$**.

The second surface of this lens is flat, so **$R_2 = \infty$**.

Applying the Lens Maker's formula again:

$$\frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{-R} - \frac{1}{\infty}\right)$$

This simplifies to:

$$\frac{1}{f_2} = \frac{-(\mu_2 - 1)}{R}$$

Notice the negative sign, which perfectly aligns with the fact that a plano-concave lens is inherently diverging.

### Combining the Powers

With the individual powers calculated, we simply add them together to find the equivalent power of the system:

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

$$\frac{1}{f_{eq}} = \frac{\mu_1 - 1}{R} + \frac{-(\mu_2 - 1)}{R}$$

Since they share a common denominator $R$, we can combine the numerators:

$$\frac{1}{f_{eq}} = \frac{\mu_1 - 1 - \mu_2 + 1}{R}$$

The $-1$ and $+1$ cancel each other out flawlessly, leaving us with:

$$\frac{1}{f_{eq}} = \frac{\mu_1 - \mu_2}{R}$$

### The Final Reveal and Edge Cases

To find the equivalent focal length, we just take the reciprocal of our result:

**$$f_{eq} = \frac{R}{\mu_1 - \mu_2}$$**

This matches option (d). But before we finish, let's appreciate the physics hidden in this equation.

What if both lenses were made of the exact same material ($\mu_1 = \mu_2$)? The denominator becomes zero, making $f_{eq} = \infty$. The system would simply act as a flat glass slab, neither converging nor diverging light.

What if the second lens was optically denser than the first ($\mu_2 > \mu_1$)? The denominator becomes negative, resulting in a negative focal length. The entire block would behave as a diverging lens, despite having a flat exterior! Physics is truly elegant when you push the equations to their limits.

Select Answer:

Visualized Solution

Analyzing the Lens Combination

The Master Formulas

Focal Length of Lens 1 (Setup)

Focal Length of Lens 1 (Compute)

Focal Length of Lens 2 (Setup)

Focal Length of Lens 2 (Compute)

Equivalent Focal Length (Setup)

Equivalent Focal Length (Compute)

Final Answer

The Way Forward

The Sigma Insight: Lens

The Setup

The Master Equation

Analyzing the First Lens

Analyzing the Second Lens

Combining the Powers

The Final Reveal and Edge Cases

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