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:

Sign convention is the only place students make mistakes here.
Remember: A plane surface has \( R = \infty \). A standard convex lens has \( R_1 > 0, R_2 < 0 \). A standard concave lens has \( R_1 0 \).

Answer 1

The Correct Option is B

Concept:
• The focal length of an individual thin lens is determined by its physical shape (radii of curvature) and material (refractive index) using the Lens Maker's Formula: \( \frac{1}{f} = (\mu - 1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \).

• When two thin lenses with focal lengths \( f_1 \) and \( f_2 \) are placed perfectly in contact, their optical powers add up.

• The equivalent focal length \( F \) of the combination is given by: \( \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \).

• Strict adherence to the Cartesian sign convention is crucial: surfaces bulging towards the incident light have positive \( R \), and surfaces cupping away have negative \( R \).

Step 1: Determine the focal length of the plano-convex lens (\( f_1 \))
Assume light travels from left to right.
For the plano-convex lens, the first surface is plane, so its radius of curvature \( R_1 = \infty \).
The second surface is convex, bulging outward to the right, so its center of curvature lies to the left (against incident light direction). Thus, \( R_2 = -R \).
Apply the Lens Maker's Formula:
\[ \frac{1}{f_1} = (\mu - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) \]
\[ \frac{1}{f_1} = (\mu - 1) \left( 0 + \frac{1}{R} \right) = \frac{\mu - 1}{R} \]

Step 2: Determine the focal length of the equi-concave lens (\( f_2 \))
For the equi-concave lens, the first surface is concave, cupping inward. Its center of curvature lies to the left, so \( R_1 = -R \).
The second surface is also concave from the perspective of exiting light, its center lies to the right, so \( R_2 = +R \).
Apply the Lens Maker's Formula:
\[ \frac{1}{f_2} = (\mu - 1) \left( \frac{1}{-R} - \frac{1}{+R} \right) \]
\[ \frac{1}{f_2} = (\mu - 1) \left( -\frac{2}{R} \right) = \frac{-2(\mu - 1)}{R} \]

Step 3: Calculate the equivalent focal length of the combination
Sum the inverse focal lengths:
\[ \frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} \]
\[ \frac{1}{F} = \frac{\mu - 1}{R} + \left( \frac{-2(\mu - 1)}{R} \right) \]
\[ \frac{1}{F} = \frac{(\mu - 1) - 2(\mu - 1)}{R} \]
\[ \frac{1}{F} = \frac{-(\mu - 1)}{R} \]
Inverting both sides gives the equivalent focal length:
\[ F = -\frac{R}{\mu - 1} \]

Step 4: Conclusion
The effective focal length of the two-lens combination is \( -\frac{R}{\mu - 1} \), which corresponds to option (B).