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JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
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
- A \(18\,cm, 20\,cm\)
- B \(10\,cm, 12\,cm\)
- C \(12\,cm, 14\,cm\)
- D \(16\,cm, 18\,cm\)
Answer & Solution
Correct Answer
(A) \(18\,cm, 20\,cm\)
Step-by-step Solution
Detailed explanation
For minimum spherical aberration separation \(d \,= f_1-f_2\, = 2\,cm\) Resultant focal lenght \(=F\,= 10\,cm\) Using \(\frac{1}{F} = \frac{1}{{{f_1}}} + \frac{1}{{{f_2}}} - \frac{d}{{{f_1}{f_2}}}\) and solving we get \(f_1\), \(f_2\) \(18\,cm\) and \(20\,cm\) respectively
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