    Question

# A plano convex lens (focal length f2, refractive index μ2 , radius of curvature R) fits exactly into a plano-concave lens (focal length f1, refractive index μ1, radius of curvature R. Their plane surfaces are parallel to each other. Then, the focal length of the combination will be

A

f1f2

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B

Rμ2μ1

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C

f1+f2

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D
2f1f2f1+f2
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Solution

## The correct option is B Rμ2−μ1 Given combination is as shown below As lenses are in contact, equivalent focal length of combination is 1feq=1f1+1f2 Using lens Maker's formula, 1f=(μ−1)(1R1−1R2) Here, 1f1=(μ1−1)(1−R−1∞)=(1−μ1)R and 1f2=(μ2−1)(1∞−1(−R))=(μ2−1)R ∴ 1feq=(μ2−1R)+(1−μ1R) =μ2−1+1−μ1R=μ2−μ1R So, feq=Rμ2−μ1 Why this question? Importance in JEE: The questions based on Lens Maker's formula, 1f=(μ−1)(1R1−1R2) are common in JEE and other competitive exams.  Suggest Corrections  7      Similar questions
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