Ray Optics

JEE Main level, single correct, medium

192 of 295

Question 192 of 295

One plano-convex and one plano-concave lens of the same radius of curvature RR but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 11 is μ1{\mu }_{1} and that of 22 isμ2{\mu }_{2}, then the focal length of the combination is:

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Answer: (D) Rμ1−μ2\dfrac{R}{{\mu }_{1}-{\mu }_{2}}

From lens maker’s formula,
1f=(μ−1)(1R1−1R2)\dfrac{1}{f}=\left(\mu -1\right)\left(\dfrac{1}{{R}_{1}}-\dfrac{1}{{R}_{2}}\right)
1f1=(μ1−1)(1∞−1(−R))=μ1−1R\dfrac{1}{{f}_{1}}=\left({\mu }_{1}-1\right)\left(\dfrac{1}{\infty }-\dfrac{1}{\left(-R\right)}\right)=\dfrac{{\mu }_{1}-1}{R}
1f2=(μ2−1)(1(−R)−1∞)=−(μ2−1)R\dfrac{1}{{f}_{2}}=\left({\mu }_{2}-1\right)\left(\dfrac{1}{\left(-R\right)}-\dfrac{1}{\infty }\right)=-\dfrac{\left({\mu }_{2}-1\right)}{R}
1feq=1f1+1f2=μ1−1R−μ2−1R\dfrac{1}{{f}_{eq}}=\dfrac{1}{{f}_{1}}+\dfrac{1}{{f}_{2}}=\dfrac{{\mu }_{1}-1}{R}-\dfrac{{\mu }_{2}-1}{R}
=μ1−μ2R=\dfrac{{\mu }_{1}-{\mu }_{2}}{R}
feq=Rμ1−μ2{f}_{eq}=\dfrac{R}{{\mu }_{1}-{\mu }_{2}}

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