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)

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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