A thin convex lens made from crown glass (… - JEE Main 2014 Physics PYQ

JEE Main 2014 — Physics Optics

2014mcqmedium

A thin convex lens made from crown glass $( \mu = \frac{ 3 }{ 2 } )$ has focal length $f⁡$ . When it is measured in two different liquids having refractive indices $\frac{ 4 }{ 3 }$ and $\frac{ 5 }{ 3 }$ , it has the focal lengths ${ f⁡ }_{1}$ and ${ f⁡ }_{2}$ respectively. The correct relation between the focal lengths is :

Official previous-year question

Held on 6 Apr 2014 · Verified 6 Jul 2026.

Options

A
${ f }_{1} = { f }_{2} < f$
B
${ f }_{1} > f$ and ${ f }_{2}$ becomes negative
C
${ f }_{2} > f$ and ${ f }_{1}$ becomes negative
D
${ f }_{1}$ and ${ f }_{2}$ both become negative

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Solution

$\text{Let} [ \frac{ 1 }{ {\text{R}}_{1} } - \frac{ 1 }{ {\text{R}}_{2} } ] = \text{x}$

$\frac{ 1 }{ \text{f} } = [ \frac{ \frac{ 3 }{ 2 } }{ 1 } - 1 ] [ \frac{ 1 }{ {\text{R}}_{1} } - \frac{ 1 }{ {\text{R}}_{2} } ] \Rightarrow \text{f} = \frac{ 2 }{ \text{x} }$

When in liquid 1

$\frac{ 1 }{ {\text{f}}_{1} } = [ \frac{ \frac{ 3 }{ 2 } }{ \frac{ 4 }{ 3 } } - 1 ] [ \frac{ 1 }{ {\text{R}}_{1} } - \frac{ 1 }{ {\text{R}}_{2} } ] \Rightarrow {\text{f}}_{1} = \frac{ 8 }{ \text{x} }$

When in liquid 2

$\frac{ 1 }{ {\text{f}}_{2} } = [ \frac{ \frac{ 3 }{ 2 } }{ \frac{ 5 }{ 3 } } - 1 ] [ \frac{ 1 }{ {\text{R}}_{1} } - \frac{ 1 }{ {\text{R}}_{2} } ] \Rightarrow {\text{f}}_{2} = \frac{ - \text{10} }{ \text{x} }$

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