Q68 / 90 · JEE Main 2019 (12 Jan Shift 2) · Physics

A plano - convex lens (focal length $f_{2}$ , refractive index $μ_{2},$ radius of curvature R) fits exactly into a plano - concave lens (focal length $f_{1},$ refractive index $μ_{1},$ radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be:

Solution

\frac{1}{f_{1}} = (μ_{1} - 1) (\frac{1}{- R} - \frac{1}{ω}) \frac{1}{f_{2}} = (μ_{2} - 1) (\frac{1}{R} - \frac{1}{ω})

When joined together

\frac{1}{f_{e q}} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{μ_{2} - 1}{- R} + \frac{μ_{2} - 1}{R} = \frac{μ_{2} - μ_{1}}{R}

\Rightarrow f_{e q} = \frac{R}{μ_{2} - μ_{1}}

Hence, the correct answer is option: 'A'.