Curved surfaces of a plano-convex lens of refractive index $\mu_{1}$ and a planoconcave lens of refractive index $\mu_{2}$ have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

(A) $\frac{1}{\mu_{2}-\mu_{1}}$

(C) $\frac{1}{\mu_{1}-\mu_{2}}$

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Geometrical Optics

Curved surfaces of a plano-convex lens of refractive index

μ_{1}

and a planoconcave lens of refractive index

μ_{2}

have equal radius of curvature as shown in figure. Find the ratio of radius of curvature to the focal length of the combined lenses.

Answer: (D)

Each surface contributes through the lens-maker's formula. The plano-convex part of index

μ_{1}

has a flat face and a convex face of radius

R

, giving

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

. The plano-concave part of index

μ_{2}

shares that curved interface (radius

R

, opposite curvature) and a flat outer face, giving

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

. In contact the powers add:

\frac{1}{f} = \frac{1}{f _{1}} + \frac{1}{f _{2}} = \frac{( μ _{1} - 1 ) - ( μ _{2} - 1 )}{R} = \frac{μ _{1} - μ _{2}}{R} .

Therefore

\frac{R}{f} = μ_{1} - μ_{2}

JEE Main · 2021 · 27 Aug 2021 · Shift 2 · Single correct option