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  • Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is

Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index 1.7. The focal length of the combination is:

Option: 1

-50 cm


Option: 2

50 cm


Option: 3

-20 cm


Option: 4

-25 cm


Answers (1)

As we learnt in

Lens placed close to each other -

frac{1}{f_{eq}}= frac{1}{f_{1}}+ frac{1}{f_{2}}------+ frac{1}{f_{n}}

frac{1}{f_{eq}}=Equivalent focal length.

 

- wherein

f_{1},f_{2},------f_{n} are focal lenght of lenc 1, 2, 3, -----n

 

 

 

 

 

Equivalent focal lenght

\frac{1}{F}= \frac{1}{F_1}+\frac{1}{F_2}+\frac{1}{F_3}

\frac{1}{F_1}=(1.5-1)(\frac{1}{\infty }-\frac{1}{-20}) =\frac{1}{40}

\frac{1}{F_2}=(1.7-1)(\frac{-1}{20 }-\frac{1}{20}) =\frac{-1.4}{20}

\frac{1}{F_3}=(1.5-1)(\frac{1}{20 }-\frac{1}{\infty }) =\frac{1}{40}

\frac{1}{F_{eq}}=\frac{1}{40}-\frac{2.8}{40}+\frac{1}{40}=-\frac{0.8}{40}

{F_{eq}}=-50 cm

Posted by

Ramraj Saini

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