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A thin convex lens made from crown glass $\left ( \mu =\frac{3}{2} \right )$ 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 :

Option 1)
$f_{1}=f_{2}<f$

Option 2)
$f_{1}>f$ and $f_{2}$ becomes negative

Option 3)
$f_{2}>f$ and $f_{1}$ becomes negative

Option 4)
$f_{1}\: and\: f_{2}$ both becomes negative

Answers (1)

best_answer

As we learnt in

Lensmaker's Formula -

$\frac{1}{f}= \left ( \frac{\mu _{2}}{\mu _{1}}-1 \right )\left ( \frac{1}{R_{1}}- \frac{1}{R_{2}}\right )$

- wherein

$\mu _{1}=$ refractive index of medium of object

$\mu _{2}=$ refractive index of lens

$R_{1}\, and \, R_{2}$ are radius of curvature of two surface

$\frac{f_{m}}{f}=\frac{(\mu-1)}{\left(\frac{\mu}{\mu_{m} }-1 \right )}$

$\frac{f_{1}}{f}=\frac{(3/2-1)}{\left(\frac{3/2}{5/3_{m} }-1 \right )}=4\ \Rightarrow\ \;f_{2}=-5f$

$\therefore\ \;f_{2}<0$

Now, f₁ > f and f₂ become negative.

Correct option is 2.

Option 1)

$f_{1}=f_{2}<f$

This is an incorrect option.

Option 2)

$f_{1}>f$ and $f_{2}$ becomes negative

This is the correct option.

Option 3)

$f_{2}>f$ and $f_{1}$ becomes negative

This is an incorrect option.

Option 4)

$f_{1}\: and\: f_{2}$ both becomes negative

This is an incorrect option.

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