Two identical thin Plano-convex glass lenses (refractive index 1 .5 ) each having radius of curvature of 20…

1. $-20\mathrm{ }\mathrm{c}\mathrm{m}$
2. $-25\mathrm{ }\mathrm{c}\mathrm{m}$
3. $-50\mathrm{ }\mathrm{c}\mathrm{m}$
4. $50\mathrm{ }\mathrm{c}\mathrm{m}$

Solution

Len's maker's formula
$\frac{1}{f} = \left(\mu -1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$
$\frac{1}{f_1}=\left(\mu -1\right)\frac{1}{ 20} = \left(1.5-1\right) \frac{1}{20}=\frac{1}{40}$
$\frac{1}{f_1}=\frac{1}{40}$
For lens $\frac{1}{f_1}+\frac{1}{f_2} = \frac{1}{40}+\frac{1}{40}=\frac{1}{20}$
For concave $\frac{1}{f_3}=\left(1.7-1\right) (-\frac{1}{R}-\frac{1}{R} )= \frac{-0.7\times 2}{R}$
$\frac{1}{f_{eq}}=\frac{1}{20}+ \left(\frac{-0.7\times 2}{R}\right)$
$=\frac{1}{20} - \frac{1.4}{20}$
$f_{eq}= -\frac{20}{0.4} =-50 cm$

Asked in: NEET 2015 (Phase 1)

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