Question

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

• A. \(\dfrac{R}{\mu_1-\mu_2}\)
• B. \(\dfrac{2R}{\mu_1-\mu_2}\)
• C. \(\dfrac{R}{2(\mu_1-\mu_2)}\)
• D. \(\dfrac{R}{2-(\mu_1+\mu_2)}\)

Hint:

For lenses in contact: \[ P_{\text{net}}=\sum P \] Sign convention is crucial.

Answer 1

The Correct Option is A

Step 1: Power of plano-convex lens: \[ P_1=\frac{\mu_1-1}{R} \]
Step 2: Power of plano-concave lens: \[ P_2=-\frac{\mu_2-1}{R} \]
Step 3: Net power: \[ P=P_1+P_2=\frac{\mu_1-\mu_2}{R} \]
Step 4: Focal length: \[ f=\frac{1}{P}=\frac{R}{\mu_1-\mu_2} \]