Given a thin convex lens (refractive index μ2), kept in a liquid (refractive index μ1, μ1 < μ2) having radii of curvature |R1| and |R2|. 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?

μ 1 R 1 ⋅ R 2 μ 2 R 1 + R 2 − μ 1 R 2

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

μ 1 R 1 ⋅ R 2 μ 2 R 1 + R 2 − μ 1 R 2

μ 1 R 1 ⋅ R 2 μ 2 2 R 1 + R 2 − μ 1 R 1 | ⋅ | R 2 ∣

μ 1 R 1 ⋅ R 2 μ 2 R 1 + R 2 − μ 1 R 2

μ 1 R 1 ⋅ R 2 μ 2 R 1 + R 2 − μ 1 R 1

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Given a thin convex lens (refractive index $μ$ ₂), kept in a liquid (refractive index $μ$ ₁, $μ$ ₁ < $μ$ ₂) having radii of curvature |R₁| and |R₂|. 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?

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$\frac{μ_{1} |R_{1}| \cdot |R_{2}|}{μ_{2} (2 |R_{1}| + |R_{2}|) - μ_{1} \sqrt{R_{1} | \cdot | R_{2} ∣}}$

$\frac{μ_{1} |R_{1}| \cdot |R_{2}|}{μ_{2} (|R_{1}| + |R_{2}|) - μ_{1} |R_{2}|}$

$\frac{μ_{1} |R_{1}| \cdot |R_{2}|}{μ_{2} (|R_{1}| + |R_{2}|) - μ_{1} |R_{1}|}$

$\frac{(μ_{2} + μ_{1}) |R_{1}|}{(μ_{2} - μ_{1})}$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{1}{f_{eq}} = \frac{2}{f_{L}} - \frac{1}{f_{m}} \\ f_{m} = - \frac{|R_{2}|}{2} \\ \frac{1}{f_{L}} = (\frac{μ_{2}}{μ_{1}} - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}}) \end{array}$
$\begin{array}{l} \frac{1}{f_{eq}} = 2 (\frac{μ_{2} - μ_{1}}{μ_{1}}) (\frac{R_{1} + R_{2}}{R_{1} R_{2}}) + \frac{2}{R_{2}} \\ = \frac{2}{R_{2}} [\frac{(μ_{2} - μ_{1}) (R_{1} + R_{2}) + μ_{1} R_{1}}{μ_{1} R_{1}}] \\ = \frac{2}{R_{2}} [\frac{μ_{2} R_{1} + μ_{2} R_{2} - μ_{1} R_{1} - μ_{1} R_{2} + μ_{1} R_{1}}{μ_{1} R_{1}}] \\ \frac{1}{f_{eq}} = \frac{2 [μ_{2} R_{1} + μ_{2} R_{2} - μ_{1} R_{2}]}{μ_{1} R_{1} R_{2}} \end{array}$
For same size of image
u = 2f
$u = \frac{μ_{1} R_{1} R_{2}}{μ_{2} R_{1} + μ_{2} R_{2} - μ_{1} R_{2}}$

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