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JEE Advanced · Physics · 24. Ray Optics

A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index \(n\) of the first lens is \(1.5\) and that of the second lens is \(1.2\). Both the curved surface are of the same radius of curvature \(R=14\) \(\mathrm{cm}\). For this bi-convex lens, for an object distance of 40 \(\mathrm{cm}\), the image distance will be

  1. A \(-280.0 \mathrm{~cm}\)
  2. B \(40.0 \mathrm{~cm}\)
  3. C \(21.5 \mathrm{~cm}\)
  4. D \(13.3 \mathrm{~cm}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(40.0 \mathrm{~cm}\)

Step-by-step Solution

Detailed explanation

The focal length \(\left(f_{1}\right)\) of the plano-cöneex Tens with \(n=1.5\) using lens-maker formula
\(\frac{1}{f_{1}}=\left(n_{1}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]=(1.5-1)\)\(\left[\frac{1}{14}-\frac{1}{\infty}\right]=\frac{1}{28}\)
The focal length \(\left(f_{2}\right)\) of the plano-convex lens with \(n=1.2\)
\(\frac{1}{f_{2}}=\left(n_{2}-1\right)\left[\frac{1}{R_{1}}-\frac{1}{R_{2}}\right]=(1.2-1)\)\(\left[\frac{1}{\infty}-\frac{1}{-14}\right]=\frac{1}{70}\)
Focal length F of the combination
\(\frac{1}{F}=\frac{1}{f_{1}}+\frac{1}{f_{2}}=\frac{1}{20}\)
Now, applying lens formula for the combination of lens
\(\frac{1}{V}-\frac{1}{U}=\frac{1}{F} \Rightarrow \frac{1}{V}-\frac{1}{-40}=\frac{1}{20}\)\([\text { Given } \mu=40 \mathrm{~cm}]\)
\(\therefore V=40 \mathrm{~cm}\)
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