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

10 c m, 12 c m

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12 c m, 14 c m

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16 c m, 18 c m

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18 c m, 20 c m

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Solution

The correct option is C $18 c m, 20 c m$
Lens placed close to each other have focal length as :

$\frac{1}{f_{e q}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + . . . + \frac{1}{f_{n}}$

$\frac{1}{10} = \frac{f_{1} + f_{2} - 2}{f_{1} f_{2}}$

$f_{1} f_{2} - 2 = \frac{f_{1} f_{2}}{10}$

So, 18cm, 20cm is correct option as it satisfies the above equation.

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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 :

The correct option is C 18 cm,20 cmLens placed close to each other have focal length as :1feq=1f1+1f2+...+1fn110=f1+f2−2f1f2f1f2−2=f1f210So, 18cm, 20cm is correct option as it satisfies the above equation.

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