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

(B) $\frac{\mathrm{R}}{\mu_{2}-\mu_{1}}$

(C) $\frac{2 f_{1} f_{2}}{f_{1}+f_{2}}$

(D) $\mathrm{f}_{1}+\mathrm{f}_{2}$

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Geometrical Optics

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 :

Answer: (B)

The plano-convex lens has one curved face of radius

R

, so

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

; the plano-concave lens has a curved face of radius

R

of the opposite sense, so

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

. Placed in contact their powers add:

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

. Hence the focal length of the pair is

f = \frac{R}{μ _{2} - μ _{1}}

JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Single correct option