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: Visualise the combination.
A plano-convex lens of index $\mu_1$ fits snugly into a plano-concave lens of index $\mu_2$ along a shared curved surface of radius $R$, with their flat faces outward. We want the focal length of the pair.
Step 2: Tool of choice.
Use the lens maker's formula $\dfrac{1}{f} = (\mu - 1)\left(\dfrac{1}{R_1} - \dfrac{1}{R_2}\right)$ for each lens, then add powers for lenses in contact.
Step 3: First lens (plano-convex).
Flat front means $R_1 = \infty$, curved back gives $R_2 = -R$: \[ \frac{1}{f_1} = (\mu_1 - 1)\left(0 + \frac{1}{R}\right) = \frac{\mu_1 - 1}{R} \]
Step 4: Second lens (plano-concave).
Its curved front matches the boundary, $R_1 = -R$, and flat back $R_2 = \infty$: \[ \frac{1}{f_2} = (\mu_2 - 1)\left(-\frac{1}{R} - 0\right) = -\frac{\mu_2 - 1}{R} \]
Step 5: Add the powers.
\[ \frac{1}{f} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R} = \frac{(\mu_1 - 1) - (\mu_2 - 1)}{R} = \frac{\mu_1 - \mu_2}{R} \]
Step 6: Invert for focal length.
$f = \dfrac{R}{\mu_1 - \mu_2}$.
\[ \boxed{f = \dfrac{R}{\mu_1 - \mu_2}\ \text{(option 1)}} \]