Question

Three lenses are arranged as shown. Find the power of equivalent lens in terms of \(R_1 \ \&\ R_2\)

(1) \(-\frac{1}{3}\left( \frac{1}{|R_1| } - \frac{1}{|R_2|}\right)\)

(2) \(\frac{1}{3} \left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right)\)

(3) \( - \frac{1}{6}\left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right)\)

(4) \(\frac{1}{6}\left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right)\)

Answer 1

Correct option is : (3) \( - \frac{1}{6}\left(\frac{1}{|R_1|} - \frac{1}{|R_2|}\right)\) 

\(\frac{1}{\mathrm{f}_{1}}=\left(\frac{4}{3}-1\right)\left(\frac{1}{\infty}+\frac{1}{\mathrm{R}_{1}}\right)=\frac{1}{3 \mathrm{R}_{1}}\)

\( \frac{1}{\mathrm{f}_{2}}=\left(\frac{3}{2}-1\right)\left(\frac{-1}{\mathrm{R}_{1}}+\frac{1}{\mathrm{R}_{2}}\right)=\frac{1}{2 \mathrm{R}_{2}}-\frac{1}{2 \mathrm{R}_{1}}\)

\( \frac{1}{\mathrm{f}_{3}}=\left(\frac{4}{3}-1\right)\left(\frac{-1}{\mathrm{R}_{2}}+\frac{1}{\infty}\right)=\frac{-1}{3 \mathrm{R}_{2}}\) 

\( \mathrm{P}_{\mathrm{eq}}=\frac{1}{3 \mathrm{R}_{1}}+\frac{1}{2 \mathrm{R}_{2}}-\frac{1}{2 \mathrm{R}_{1}}-\frac{1}{3 \mathrm{R}_{2}}\)

\( P_{e q}=\frac{-1}{6 R_{1}}+\frac{1}{6 R_{2}}\)