Q147 / 180 · COMEDK 2025 (25 May Shift) · Physics
A plano-convex lens made of refractive index 1.5 and having radius of curvature $\mathrm{R}=4 \mathrm{~cm}$ fits exactly into a plano-concave lens made of refractive index 1.3 and having the same radius of curvature $\mathrm{R}=4 \mathrm{~cm}$ such that their plane surfaces are parallel to each other. The focal length of the combination is:
- A5 cm
- B2 cm
- C50 cm
- Correct answerD20 cm
Solution
For the plano-convex lens ($\mu_1 = 1.5$), $R_1 = \infty$, $R_2 = -4$ cm:
$\dfrac{1}{f_1} = (1.5 - 1)\left(\dfrac{1}{\infty} - \dfrac{1}{-4}\right) = 0.5 \times \dfrac{1}{4} = \dfrac{0.5}{4}$ cm$^{-1}$
For the plano-concave lens ($\mu_2 = 1.3$), $R_1 = -4$ cm, $R_2 = \infty$:
$\dfrac{1}{f_2} = (1.3 - 1)\left(\dfrac{1}{-4} - \dfrac{1}{\infty}\right) = 0.3 \times \left(-\dfrac{1}{4}\right) = -\dfrac{0.3}{4}$ cm$^{-1}$
Equivalent focal length:
$\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2} = \dfrac{0.5}{4} - \dfrac{0.3}{4} = \dfrac{0.2}{4} = \dfrac{1}{20}$ cm$^{-1}$
$F = 20$ cm
$\dfrac{1}{f_1} = (1.5 - 1)\left(\dfrac{1}{\infty} - \dfrac{1}{-4}\right) = 0.5 \times \dfrac{1}{4} = \dfrac{0.5}{4}$ cm$^{-1}$
For the plano-concave lens ($\mu_2 = 1.3$), $R_1 = -4$ cm, $R_2 = \infty$:
$\dfrac{1}{f_2} = (1.3 - 1)\left(\dfrac{1}{-4} - \dfrac{1}{\infty}\right) = 0.3 \times \left(-\dfrac{1}{4}\right) = -\dfrac{0.3}{4}$ cm$^{-1}$
Equivalent focal length:
$\dfrac{1}{F} = \dfrac{1}{f_1} + \dfrac{1}{f_2} = \dfrac{0.5}{4} - \dfrac{0.3}{4} = \dfrac{0.2}{4} = \dfrac{1}{20}$ cm$^{-1}$
$F = 20$ cm
Hence, the correct answer is option: 'D'.