Ray Optics

JEE Main level, single correct, medium

195 of 295

Question 195 of 295

A plano - convex lens (focal length f2{f}_{2} , refractive index μ2,{\mu }_{2}, radius of curvature R) fits exactly into a plano - concave lens (focal length f1,{f}_{1}, refractive index μ1,{\mu }_{1}, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be:

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Answer: (A) Rμ2−μ1\dfrac{R}{{\mu }_{2}-{\mu }_{1}}

1f1=(μ1−1)(1−R−1ω)1f2=(μ2−1)(1R−1ω)\dfrac{1}{{f}_{1}}=\left({\mu }_{1}-1\right)\left(\dfrac{1}{-R}-\dfrac{1}{\omega }\right)\dfrac{1}{{f}_{2}}=\left({\mu }_{2}-1\right)\left(\dfrac{1}{R}-\dfrac{1}{\omega }\right)

When joined together

1feq=1f1+1f2=μ2−1−R+μ2−1R=μ2−μ1R\dfrac{1}{{f}_{eq}}=\dfrac{1}{{f}_{1}}+\dfrac{1}{{f}_{2}}=\dfrac{{\mu }_{2}-1}{-R}+\dfrac{{\mu }_{2}-1}{R}=\dfrac{{\mu }_{2}-{\mu }_{1}}{R}

⇒feq=Rμ2−μ1\Rightarrow {f}_{eq}=\dfrac{R}{{\mu }_{2}-{\mu }_{1}}

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