Question 151 of 295

Given a thin convex lens (refractive index $\mu_2$ ), kept in a liquid (refractive index $\mu_1, \mu_1 \lt \mu_2$ ) having radii of curvatures $\left|R_1\right|$ and $\left|R_2\right|$. 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?

A$\frac{\mu_1\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}{\mu_2\left(\left|\mathrm{R}_1\right|+\left|\mathrm{R}_2\right|\right)-\mu_1\left|\mathrm{R}_2\right|}$B$\frac{\mu_1\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}{\mu_2\left(\left|\mathrm{R}_1\right|+\left|\mathrm{R}_2\right|\right)-\mu_1\left|\mathrm{R}_1\right|}$C$\frac{\left(\mu_2+\mu_1\right)\left|\mathrm{R}_1\right|}{\left(\mu_2-\mu_1\right)}$D$\frac{\mu_1\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}{\mu_2\left(2\left|\mathrm{R}_1\right|+\left|\mathrm{R}_2\right|\right)-\mu_1 \sqrt{\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}}$

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Answer: (A) $\frac{\mu_1\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}{\mu_2\left(\left|\mathrm{R}_1\right|+\left|\mathrm{R}_2\right|\right)-\mu_1\left|\mathrm{R}_2\right|}$

$\begin{aligned} & \frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{2}{\mathrm{f}_{\mathrm{L}}}-\frac{1}{\mathrm{f}_{\mathrm{m}}} \\ & \mathrm{f}_{\mathrm{m}}=-\frac{\left|\mathrm{R}_2\right|}{2} \\ & \frac{1}{\mathrm{f}_{\mathrm{L}}}=\left(\frac{\mu_2}{\mu_1}-1\right)\left(\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}\right)\end{aligned}$

$\begin{aligned}
& \frac{1}{\mathrm{f}_{\mathrm{eq}}}=2\left(\frac{\mu_2-\mu_1}{\mu_1}\right)\left(\frac{\mathrm{R}_1+\mathrm{R}_2}{\mathrm{R}_1 \mathrm{R}_2}\right)+\frac{2}{\mathrm{R}_2} \\
& =\frac{2}{\mathrm{R}_2}\left[\frac{\left(\mu_2-\mu_1\right)\left(\mathrm{R}_1+\mathrm{R}_2\right)+\mu_1 \mathrm{R}_1}{\mu_1 \mathrm{R}_1}\right] \\
& =\frac{2}{\mathrm{R}_2}\left[\frac{\mu_2 \mathrm{R}_1+\mu_2 \mathrm{R}_2-\mu_1 \mathrm{R}_1-\mu_1 \mathrm{R}_2+\mu_1 \mathrm{R}_1}{\mu_1 \mathrm{R}_1}\right] \\
& \frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{2\left[\mu_2 \mathrm{R}_1+\mu_2 \mathrm{R}_2-\mu_1 \mathrm{R}_2\right]}{\mu_1 \mathrm{R}_1 \mathrm{R}_2}
\end{aligned}$

For same size of image
$\begin{aligned}
& \mathrm{u}=2 \mathrm{f} \\
& \mathrm{u}=\frac{\mu_1 \mathrm{R}_1 \mathrm{R}_2}{\mu_2 \mathrm{R}_1+\mu_2 \mathrm{R}_2-\mu_1 \mathrm{R}_2}
\end{aligned}$