Step 1 (Formula / Governing Law & Given Data): [0.5 Mark]
The focal length of a thin lens is given by the Lens Maker's Formula:
$$\frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right)$$For the combination of thin lenses in contact, the equivalent focal length $F$ is given by:
$$\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2}$$Given: Refractive index = $\mu$, Radius of curvature = $R$.
Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]
- For the plano-convex lens ($f_1$): $R_1 = R$ and $R_2 = \infty$. $$\frac{1}{f_1} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) = \frac{\mu - 1}{R}$$
- For the equi-concave lens ($f_2$): $R_1 = -R$ and $R_2 = +R$. $$\frac{1}{f_2} = (\mu - 1) \left( \frac{1}{-R} - \frac{1}{R} \right) = (\mu - 1) \left( -\frac{2}{R} \right) = -\frac{2(\mu - 1)}{R}$$
- Equivalent focal length ($F$): $$\frac{1}{F} = \frac{\mu - 1}{R} - \frac{2(\mu - 1)}{R} = \frac{\mu - 1 - 2\mu + 2}{R} = \frac{1 - \mu}{R} = -\frac{\mu - 1}{R}$$
Step 3 (Final Concluded Answer): [0.25 Mark]
Taking the reciprocal of the expression above:
$$F = -\frac{R}{\mu - 1}$$Correct Option: (B)
💡 EL10 Concept Insight:
Think of the plano-convex lens as a "converger" and the equi-concave lens as a "diverger." Because the concave lens is curved on both sides, it spreads light out twice as strongly as the single-sided curve of the plano-convex lens, causing the overall combination to act like a net diverger (negative focal length).