Given a thin convex lens (refractive index μ2), kept in a liquid (refractive index μ1, μ1 < μ2)

Question

Given a thin convex lens (refractive index \(\mu_2\)), kept in a liquid (refractive index \(\mu_1,\ \mu_1<\mu_2\)) having radii of curvature \(|R_1|\ \text{and}\ |R_2|.\) Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place ?

(1) \(\frac{\mu_1|R_1|.|R_2|}{\mu_2(|R_1| + |R_2|) - \mu_1|R_1|}\)

(2) \(\frac{\mu_1|R_1|.|R_2|}{\mu_2(|R_1| + |R_2|) - \mu_1|R_2|}\)

(3) \(\frac{\mu_1|R_1|.|R_2|}{\mu_2(2|R_1| + |R_2|) - \mu_1\sqrt{|R_1|.|R_2|}}\) 

(4) \(\frac{(\mu_2 + \mu_1)|R_1|}{(\mu_2 - \mu_1)}\)

Answer 1

Correct option is : (2) \(\frac{\mu_1|R_1|.|R_2|}{\mu_2(|R_1| + |R_2|) - \mu_1|R_2|}\) 

\(\frac{1}{f_{eq}} = \frac{2}{f_L} -\frac{1}{f_m}\)

\(f_m = -\frac{|R_2|}{2}\)

\(\frac{1}{f_L} = \left(\frac{\mu_2}{\mu_1} - 1\right)\left(\frac{1}{R_1}+ \frac{1}{R_2}\right)\) 

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

\( =\frac{2}{\mathrm{R}_{2}}\left[\frac{\left(\mu_{2}-\mu_{1}\right)\left(\mathrm{R}_{1}+\mathrm{R}_{2}\right)+\mu_{1} \mathrm{R}_{1}}{\mu_{1} \mathrm{R}_{1}}\right]\) 

\( =\frac{2}{R_{2}}\left[\frac{\mu_{2} R_{1}+\mu_{2} R_{2}-\mu_{1} R_{1}-\mu_{1} R_{2}+\mu_{1} R_{1}}{\mu_{1} R_{1}}\right]\) 

\( \frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{2\left[\mu_{2} \mathrm{R}_{1}+\mu_{2} \mathrm{R}_{2}-\mu_{1} \mathrm{R}_{2}\right]}{\mu_{1} \mathrm{R}_{1} \mathrm{R}_{2}}\) 

For same size of image

u = 2f

\( u=\frac{\mu_{1} R_{1} R_{2}}{\mu_{2} R_{1}+\mu_{2} R_{2}-\mu_{1} R_{2}}\)