QJEEM18Phb845: JEE Main 2018 Physics - Class 12 JEEM18Phb845 Solution: A convergent doublet of separated lenses, corrected for…

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QJEEM18Phb845: JEE Main 2018 Physics - Class 12 JEEM18Phb845 Solution: A convergent doublet of separated lenses, corrected for…

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 :

Multiple Choice Options

A 10 cm, 12 cm
B 12 cm, 14 cm
C 16 cm, 18 cm
D 18 cm, 20cm

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Accepted Answer

For a convergent doublet of separated lens, we have

$\frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} - \frac{d}{f_{1} f_{2}}$ ...... (1)

where d is separation between two lens, f₁ and f₂ are focal lengths of component lenses, f is resultant focal length. Therefore, Eq. (1) becomes

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

$\Rightarrow f_{1} f_{2} = 10 f_{2} + 10 f_{1} - 20$

$\Rightarrow 10 f_{1} + 10 f_{2} - f_{1} f_{2} = + 20$

For f₁ = 18 cm and f₂ = 20 cm, the above equation satisfies.

QJEEM18Phb845: JEE Main 2018 Physics - Class 12 JEEM18Phb845 Solution: A convergent doublet of separated lenses, corrected for…

Multiple Choice Options

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