Back to Directory
NEET PHYSICSMedium

A plano-convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lenses are made of different materials of refractive indices μ1\mu_1 and μ2\mu_2 and RR is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is:

A

R2(μ1+μ2)\frac{R}{2(\mu_1+\mu_2)}

B

R2(μ1−μ2)\frac{R}{2(\mu_1-\mu_2)}

C

Rμ1−μ2\frac{R}{\mu_1-\mu_2}

D

2Rμ2−μ1\frac{2R}{\mu_2-\mu_1}

Step-by-Step Solution

  1. Focal Length of Plano-Convex Lens (f1f_1): Let's assume light enters the plano-convex lens through its plane surface. The radii of curvature are R1=∞R_1 = \infty and R2=−RR_2 = -R (using sign convention). Using the Lens Maker's formula: 1f1=(μ1−1)(1∞−1−R)=μ1−1R\frac{1}{f_1} = (\mu_1 - 1) \left( \frac{1}{\infty} - \frac{1}{-R} \right) = \frac{\mu_1 - 1}{R}.

  2. Focal Length of Plano-Concave Lens (f2f_2): The plano-concave lens fits exactly over the curved surface, so its curved surface has the same radius RR but is concave towards the incident light. The plane surfaces are parallel. Its radii are R1=−RR_1 = -R and R2=∞R_2 = \infty. 1f2=(μ2−1)(1−R−1∞)=−μ2−1R\frac{1}{f_2} = (\mu_2 - 1) \left( \frac{1}{-R} - \frac{1}{\infty} \right) = -\frac{\mu_2 - 1}{R}.

  3. Equivalent Focal Length (FF): Since the lenses are in contact, the equivalent focal length of the combination is obtained by adding their optical powers: 1F=1f1+1f2\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} 1F=μ1−1R−μ2−1R=(μ1−1)−(μ2−1)R=μ1−μ2R\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}.

Therefore, the focal length of the combination is F=Rμ1−μ2F = \frac{R}{\mu_1 - \mu_2}.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started
Solved: PHYSICS Question for NEET | Sushrut