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JEE MAIN 2025
23-01-2025 SHIFT-1
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 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?
Select the correct option:
A
$\frac{{{{\mu }_{1}}\left| {{{\text{R}}_{1}}} \right|\cdot \left| {{{\text{R}}_{2}}} \right|}}{{{{\mu }_{2}}\left( {\left| {{{\text{R}}_{1}}} \right|+\left| {{{\text{R}}_{2}}} \right|} \right)-{{\mu }_{1}}\left| {{{\text{R}}_{2}}} \right|}}$
B
$\frac{{({{\mu }_{2}}+{{\mu }_{1}})\left| {{{\text{R}}_{1}}} \right|}}{{({{\mu }_{2}}-{{\mu }_{1}})}}$
C
$\frac{{{{\mu }_{1}}\left| {{{\text{R}}_{1}}} \right|\cdot \left| {{{\text{R}}_{2}}} \right|}}{{{{\mu }_{2}}\left( {2\left| {{{\text{R}}_{1}}} \right|+\left| {{{\text{R}}_{2}}} \right|} \right)-{{\mu }_{1}}\sqrt{{\left| {{{\text{R}}_{1}}} \right|\cdot \left| {{{\text{R}}_{2}}} \right|}}}}$
D
$\frac{{{{\mu }_{1}}\left| {{{\text{R}}_{1}}} \right|\cdot \left| {{{\text{R}}_{2}}} \right|}}{{{{\mu }_{2}}\left( {\left| {{{\text{R}}_{1}}} \right|+\left| {{{\text{R}}_{2}}} \right|} \right)-{{\mu }_{1}}\left| {{{\text{R}}_{1}}} \right|}}$
✓ Correct! Well done.
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