Question

A plano-convex lens of refractive index $\mu_1$ fits exactly into a plano-concave lens of refractive index $\mu_2$. Their plane surfaces are parallel to each other. 'R' is the radius of curvature of the curved surface of the lenses. The focal length of the combination is

• A. $\frac{R}{\mu_1 - \mu_2}$
• B. $\frac{R}{2(\mu_1 + \mu_2)}$
• C. $\frac{2R}{\mu_1 - \mu_2}$
• D. $\frac{R}{2(\mu_1 - \mu_2)}$

Hint:

Think of this combined setup as a single lens boundary separating two different optical materials. The $(-1)$ terms in the standard single-lens formulas will always cancel out during addition, leaving behind a simple difference of indices: $\frac{1}{f} = \frac{\Delta \mu}{R}$.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question presents a combination of two lenses matching along an identical curved interface of radius $R$. One is plano-convex ($\mu_1$) and the other is plano-concave ($\mu_2$). We need to determine the effective combined focal length ($f$) of the system.

Step 2: Key Formula or Approach:
According to the Lens Maker's Formula, the focal length of a single lens component is:
$$\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$$ For multiple thin lenses in direct contact, their combined optical power adds up linearly:
$$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2}$$

Step 3: Detailed Explanation:
Let's analyze the configuration piece by piece using Cartesian sign conventions:
1.

Lens 1 (Plano-convex): Front surface is flat ($R_1 = \infty$), back surface is curved inward relative to light direction ($R_2 = -R$).
$$\frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{\infty} - \left(-\frac{1}{R}\right)\right) = \frac{\mu_1 - 1}{R}$$ 2.

Lens 2 (Plano-concave): Front surface matches the curved boundary ($R_1' = -R$), back surface is flat ($R_2' = \infty$).
$$\frac{1}{f_2} = (\mu_2 - 1)\left(-\frac{1}{R} - \frac{1}{\infty}\right) = -\frac{\mu_2 - 1}{R}$$ Combine the individual power values to calculate the total system focal path:
$$\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R}$$ Combine the numerators over the common denominator $R$:
$$\frac{1}{f_{\text{eq}}} = \frac{(\mu_1 - 1) - (\mu_2 - 1)}{R} = \frac{\mu_1 - 1 - \mu_2 + 1}{R} = \frac{\mu_1 - \mu_2}{R}$$ Inverting both sides to isolate the equivalent focal length $f_{\text{eq}}$:
$$f_{\text{eq}} = \frac{R}{\mu_1 - \mu_2}$$

Step 4: Final Answer:
The focal length of the combination is $\frac{R}{\mu_1 - \mu_2}$, which corresponds to option (A).