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  • Stuck here, help me understand: A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is 2 cm. The focal lengths of the component lenses are :

A convergent doublet of separated lenses, corrected for spherical aberration, has resultant focal length of 10 cm. The separation between the two lenses is 2 cm. The focal lengths of the component lenses are :

  • Option 1)

    10 cm, 12 cm

  • Option 2)

    12 cm, 14 cm

  • Option 3)

    16 cm, 18 cm

  • Option 4)

    18 cm, 20 cm

 

Answers (2)

best_answer

As we learnt

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

 

 \frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}-\frac{d}{f_{1}f_{2}}

\therefore \frac{1}{10}=\frac{f_{1}+f_{2}-2}{f_{1}f_{2}}\: \: or\: \: f_{1}+f_{2}-2=\frac{f_{1}f_{2}}{10}

(4) satisfies eqn.


Option 1)

10 cm, 12 cm

Option 2)

12 cm, 14 cm

Option 3)

16 cm, 18 cm

Option 4)

18 cm, 20 cm

Posted by

Avinash

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