Given a thin convex lens (refractive index ), kept in a liquid (refractive indes ) having radii of curvature || and ||. 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 a

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Solution

Q.
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 | $R_{1}$ | 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 [2025]

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

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

3 $\frac{μ_{1} | R_{1} | \cdot | R_{2} |}{μ_{2} (2 | R_{1} | + | R_{2} |) - μ_{1} \sqrt{| R_{1} | \cdot | R_{2} |}}$

4 $\frac{(μ_{2} + μ_{1}) | R_{1} |}{(μ_{2} - μ_{1})}$

Ans.
(2)

$\frac{1}{f_{l}} = (\frac{μ_{2}}{μ_{1}} - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}})$

$f_{m} = - \frac{R_{2}}{2}$

$P_{eq} = 2 P_{l} + P_{m}$

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

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

$\frac{1}{f} = \frac{2 μ_{2} R_{1} + 2 μ_{2} R_{2} - 2 μ_{1} R_{1} - 2 μ_{1} R_{2} + 2 μ_{1} R_{1}}{μ_{1} R_{1} R_{2}}$

$\Rightarrow f = \frac{μ_{1} R_{1} R_{2}}{2 μ_{2} R_{1} + 2 μ_{2} R_{2} - 2 μ_{1} R_{2}}$

Required distance = $2 f$

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