JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Geometrical Optics

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One plano-convex and one plano-concave lens of same radius of curvature ' R ' but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is

μ_{1}

and that of 2 is

μ_{2}

, then the focal length of the combination is

(A) $\frac{R}{2 ( μ _{1} - μ _{2} )}$
(B) $\frac{2 R}{μ _{1} - μ _{2}}$
(C) $\frac{R}{μ _{1} - μ _{2}}$ ✓ Correct
(D) $\frac{R}{2 - ( μ _{1} - μ _{2} )}$

Answer: (C)

Solution

Treating the two thin lenses in contact, the lens maker's formula gives

\frac{1}{f _{1}} = \frac{μ _{1} - 1}{R}

for the plano-convex lens and

\frac{1}{f _{2}} = - \frac{μ _{2} - 1}{R}

for the plano-concave lens of the same radius. Adding the powers,

\frac{1}{f _{e q}} = \frac{μ _{1} - 1}{R} - \frac{μ _{2} - 1}{R} = \frac{μ _{1} - μ _{2}}{R}

, so the combination behaves as a single lens of focal length

f_{e q} = \frac{R}{μ _{1} - μ _{2}}

JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 10 Apr 2019 · Shift 1 · Geometrical Optics

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