A plano-convex lens (focal length f2, refractive index mu2, radius of curvature R) fits exactly into

Q: Why is the radius of the curved surface taken as $-R$ for both lenses?

According to the Cartesian sign convention, the direction of incident light is taken as positive (usually left to right). For the plano-convex lens, the second surface is convex, meaning its center of curvature lies to the left (against the incident light), so $R_2 = -R$. For the plano-concave lens, the first surface is concave, and its center of curvature also lies to the left, so $R_1 = -R$.

Q: Does the order of the lenses matter in this combination?

No, for thin lenses in contact, the equivalent focal length $\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$ is independent of the order in which the light passes through them. The physical result remains the same.

Q: What would happen if $\mu_1 = \mu_2$?

If the refractive indices are equal, the equivalent focal length becomes infinity ($f_{eq} = \frac{R}{0} = \infty$). The combination perfectly cancels out any converging or diverging power, acting just like a simple, flat glass slab.

Q: Why is the focal length of a plane surface considered infinity?

A plane surface can be thought of as a part of a sphere with an infinitely large radius. Since curvature is the reciprocal of radius, a flat surface has zero curvature, which corresponds to $R = \infty$.

Q: Can we use the effective power formula $P = P_1 + P_2$ here?

Absolutely! The power of a lens is simply the reciprocal of its focal length ($P = \frac{1}{f}$). So, adding the powers $P_{eq} = P_1 + P_2$ is mathematically identical to the formula $\frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2}$ we used.

Question

A plano-convex lens (focal length $f_2$, refractive index $\mu_2$, radius of curvature $R$) fits exactly into a plano-concave lens (focal length $f_1$, refractive index $\mu_1$, radius of curvature $R$). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be

Options:

(A) $f_1 - f_2$

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

(C) $f_1 + f_2$

(D) $\frac{2f_1 f_2}{f_1 + f_2}$

Sigma Percentile · Accepted Answer

### The Art of Combining Lenses

Imagine you have two puzzle pieces made of glass. One is a plano-convex lens, and the other is a plano-concave lens. When you fit them together perfectly, they form a new, combined optical system.

This problem is a beautiful exercise in applying the **Lens Maker's Formula** and understanding the power of lenses in contact. Let's break it down piece by piece.

### Decoding the Plano-Convex Lens

Let's isolate the first puzzle piece: the plano-convex lens with refractive index $\mu_2$.

According to the Cartesian sign convention, we assume light travels from left to right. The light first encounters the flat, plane surface. A plane surface has no curvature, which means its radius of curvature is infinite ($R_1 = \infty$).

Next, the light hits the curved convex surface. Notice carefully that this surface bulges to the right, meaning its center of curvature lies to the left, against the direction of incident light. Therefore, its radius is negative ($R_2 = -R$).

Applying the Lens Maker's formula:
$$ \frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right) $$

Since $\frac{1}{\infty} = 0$, the equation simplifies elegantly:
$$ \frac{1}{f_2} = \frac{\mu_2 - 1}{R} $$

### Decoding the Plano-Concave Lens

Now, let's look at the second piece: the plano-concave lens with refractive index $\mu_1$.

Here, the light first encounters the concave surface. Just like before, the center of curvature for this surface lies to the left, so its radius is negative ($R_1 = -R$).

The light then exits through the plane surface, which has an infinite radius ($R_2 = \infty$).

Applying the Lens Maker's formula again:
$$ \frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{-R} - \frac{1}{\infty}\right) $$

This simplifies to:
$$ \frac{1}{f_1} = \frac{-(\mu_1 - 1)}{R} $$

### The Grand Assembly

When two thin lenses are placed in contact, their optical powers add up. Since power is the reciprocal of focal length, the equivalent focal length $f_{eq}$ of the combination is given by:
$$ \frac{1}{f_{eq}} = \frac{1}{f_1} + \frac{1}{f_2} $$

Now, we simply substitute the individual focal lengths we just derived:
$$ \frac{1}{f_{eq}} = \frac{-(\mu_1 - 1)}{R} + \frac{\mu_2 - 1}{R} $$

Since the denominators are identical, 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 perfectly, leaving us with:
$$ \frac{1}{f_{eq}} = \frac{\mu_2 - \mu_1}{R} $$

Finally, taking the reciprocal gives us the equivalent focal length of the entire system:
$$ f_{eq} = \frac{R}{\mu_2 - \mu_1} $$

This elegant result shows how the individual refractive indices and the shared radius of curvature dictate the behavior of the combined lens.

Select Answer:

Visualized Solution

Visualizing the Lens Combination

Equivalent Focal Length Formula

Lens Maker's Formula for Lens 2

Focal Length $f_{2}$

Lens Maker's Formula for Lens 1

Focal Length $f_{1}$

Substituting into Equivalent Formula

Simplifying the Expression

Final Equivalent Focal Length $f_{e q}$

Conceptual Extensions

The Sigma Insight: Lens

The Art of Combining Lenses

Decoding the Plano-Convex Lens

Decoding the Plano-Concave Lens

The Grand Assembly

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