Question

A thin plano-convex lens and a thin equi-concave lens are kept coaxially in contact as shown in the figure. Assuming both the lenses are made of glass of refractive index \(\mu\), and \(R\) is the radius of curvature of each curved surface, the focal length of the combination is :

• A. \(\frac{R}{\mu - 1}\)
• B. \(-\frac{R}{\mu - 1}\)
• C. \(\frac{2R}{\mu - 1}\)
• D. \(-\frac{2R}{\mu - 1}\)

Hint:

When combining lenses, adding their individual powers algebraically (\(P = P_1 + P_2\)) prevents reciprocal errors. A plano-convex lens has a power of \(P_1 = \frac{\mu-1}{R}\), while an equi-concave lens has a power of \(P_2 = \frac{-2(\mu-1)}{R}\). Adding them gives \(P_{\text{net}} = \frac{-(\mu-1)}{R}\), leading directly to \(F = -\frac{R}{\mu-1}\).

Answer 1

The Correct Option is B

Concept: The focal length of any individual thin lens in air can be calculated using the Lens Maker's Formula: \[ \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where \(\mu\) is the refractive index of the lens material, and \(R_1, R_2\) are the radii of curvature of the first and second refracting surfaces, respectively, adhering strictly to Cartesian sign conventions. When two thin lenses with focal lengths \(f_1\) and \(f_2\) are kept coaxially in direct contact, the effective focal length (\(F\)) of the lens combination is given by the power addition formula: \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]

Step 1: Finding the focal length of the first lens (Plano-convex lens).
For a standard plano-convex lens where light enters from the flat or curved side: Let the first surface be convex with a positive radius of curvature: \(R_1 = +R\). The second surface is plane, so its radius of curvature is infinite: \(R_2 = \infty\). Applying the Lens Maker's Formula for this first lens (\(f_1\)): \[ \frac{1}{f_1} = (\mu - 1) \left( \frac{1}{R} - \frac{1}{\infty} \right) \] Since \(\frac{1}{\infty} = 0\), this simplifies to: \[ \frac{1}{f_1} = (\mu - 1) \left( \frac{1}{R} - 0 \right) = \frac{\mu - 1}{R} \quad \cdots (1) \]

Step 2: Finding the focal length of the second lens (Equi-concave lens).
For an equi-concave lens, both bounding surfaces are concave: The first surface curves inward, so according to the direction of incident light, its radius of curvature is negative: \(R_1 = -R\). The second surface curves outward, so its radius of curvature is positive: \(R_2 = +R\). Applying the Lens Maker's Formula for this second lens (\(f_2\)): \[ \frac{1}{f_2} = (\mu - 1) \left( \frac{1}{-R} - \frac{1}{R} \right) \] Combining the terms inside the parentheses: \[ \frac{1}{f_2} = (\mu - 1) \left( -\frac{2}{R} \right) = -\frac{2(\mu - 1)}{R} \quad \cdots (2) \]

Step 3: Calculating the equivalent focal length of the combination.
Substituting equations (1) and (2) into the combination formula: \[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \] \[ \frac{1}{F} = \frac{\mu - 1}{R} + \left( -\frac{2(\mu - 1)}{R} \right) \] Since the denominators are identical, we can directly combine the numerators: \[ \frac{1}{F} = \frac{(\mu - 1) - 2(\mu - 1)}{R} = \frac{-(\mu - 1)}{R} \] Taking the reciprocal of both sides to isolate the equivalent focal length \(F\): \[ F = -\frac{R}{\mu - 1} \] The negative sign indicates that the overall combination behaves as a net diverging (concave) lens system. This matches Option (B).