Ray Optics

JEE Main level, single correct, medium

199 of 295

Question 199 of 295

A plano-convex lens of refractive index μ1{\mu }_{1} and focal length f1{f}_{1} is kept in contact with another plano-concave lens of refractive index μ2{\mu }_{2} and focal length f2.{f}_{2}. If the radius of curvature of their spherical faces is RR each and f1=2f2,{f}_{1}=2{f}_{2}, the μ1{\mu }_{1} and μ2{\mu }_{2} are related as:

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Answer: (D) 2μ1−μ2=12{\mu }_{1}-{\mu }_{2}=1

Given,
f1=2f2{f}_{1}=2{f}_{2}
Lens maker formula for the first lens is given by,
1f1=(μ1−1)(1R−1∞)...(1)\dfrac{1}{{f}_{1}}=\left({\mu }_{1}-1\right)\left(\dfrac{1}{R}-\dfrac{1}{\infty }\right)\text{...(}1\text{)}
Lens maker formula for the second lens is given by,
1f2=(μ2−1)(1R−1∞)...(2)\dfrac{1}{{f}_{2}}=\left({\mu }_{2}-1\right)\left(\dfrac{1}{R}-\dfrac{1}{\infty }\right)\text{...(}2\text{)}
After (1)(2)\dfrac{\left(1\right)}{\left(2\right)}, we get:

f2f1=μ1−1μ2−1=12\dfrac{{f}_{2}}{{f}_{1}}=\dfrac{{\mu }_{1}-1}{{\mu }_{2}-1}=\dfrac{1}{2}

⇒2μ1−2=μ2−1\Rightarrow 2{\mu }_{1}-2={\mu }_{2}-1

⇒2μ1−μ2=1\Rightarrow 2{\mu }_{1}-{\mu }_{2}=1

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