Curved surfaces of a plano-convex lens of refractive index mu_1 and a plano-concave lens of refracti

Q: Why is $R_2$ negative for the first lens?

By standard Cartesian sign convention, we measure distances from the optical center. If light travels from left to right, the center of curvature for the right-facing convex surface lies to the left of the pole. Since we measure against the direction of incident light, the radius is taken as negative.

Q: Does it matter which side the light enters from?

No, the focal length of a thin lens is the same regardless of which side the light enters. If you assume light enters from the right, the signs of $R_1$ and $R_2$ will flip, but the final result for $\frac{1}{f}$ will remain exactly the same.

Q: Why do the $-1$ and $+1$ cancel out?

The $-1$ in the Lens Maker's formula comes from the refractive index of the surrounding medium (air, $\mu=1$). When you add the powers of the two lenses, the geometric structure of the shared boundary causes these constant terms to perfectly offset each other.

Q: What if the lenses had different radii of curvature?

If they had different radii, say $R_a$ and $R_b$, they wouldn't fit perfectly together. You would have to calculate $\frac{1}{f_1}$ using $R_a$ and $\frac{1}{f_2}$ using $R_b$, and the final expression would not simplify as cleanly.

Q: Is the combined lens converging or diverging?

It depends entirely on the refractive indices! If $\mu_1 > \mu_2$, the power is positive, making it a converging system. If $\mu_2 > \mu_1$, the power is negative, making it a diverging system.

Question

Curved surfaces of a plano-convex lens of refractive index $\mu_1$ and a plano-concave lens of refractive index $\mu_2$ have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

Options:

(A) $\frac{1}{\mu_2 - \mu_1}$

(B) $\mu_1 - \mu_2$

(C) $\frac{1}{\mu_1 - \mu_2}$

(D) $\mu_2 - \mu_1$

Sigma Percentile · Accepted Answer

### The Puzzle Pieces

Imagine you have two lenses perfectly glued together. On the left, we have a plano-convex lens with a refractive index of $\mu_1$. On the right, a plano-concave lens with a refractive index of $\mu_2$. They fit together like puzzle pieces because their curved surfaces share the exact same radius of curvature, $R$.

Our goal is to find the ratio of this radius of curvature $R$ to the equivalent focal length of the combined system. To do this, we need to break the system down and analyze each lens individually before bringing them back together.

### The Lens Maker's Magic

To find the focal length of any thin lens, we rely on our trusty Lens Maker's formula:

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

Let's isolate the first lens, the plano-convex one. Assuming light travels from left to right, it hits the flat surface first. A flat surface has an infinite radius of curvature, so $R_1 = \infty$. Next, the light hits the curved surface. Since the center of this curve lies to the left (against the direction of incoming light), we must use the Cartesian sign convention, making $R_2 = -R$.

Substituting these into the formula gives us:

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

Since $\frac{1}{\infty}$ is zero, and the two negative signs cancel out, the power of the first lens simplifies beautifully to:

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

Now, let's look at the second lens, the plano-concave one. Here, the light first encounters the curved surface. Again, its center is to the left, so $R_1 = -R$. The second surface is flat, making $R_2 = \infty$. Plugging these in:

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

This leaves us with a negative sign, which perfectly aligns with the fact that a concave surface is diverging:

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

### Bringing It All Together

When two thin lenses are in contact, their equivalent optical power is simply the sum of their individual powers.

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

Substituting the expressions we just derived:

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

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

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

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

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

The question asks for the ratio of $R$ to the focal length. By simply cross-multiplying, we find our final, elegant result:

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

Animated Solution for Physics - Optics: Curved surfaces of a plano-convex lens of refractive index $μ_{1}$ and a plano-concave lens of refractive index $μ_{2}$ have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

Select Answer:

Visualized Solution

Visual Anchor

Logic Bridge

Lens 1 Setup

Lens 1 Compute

Lens 2 Setup

Lens 2 Compute

Combine

Final Compute

The Way Forward

The Sigma Insight: Lens

The Puzzle Pieces

The Lens Maker's Magic

Bringing It All Together

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