A plano-convex lens (focal length f2, refractive index μ2, radius of curvature R) fits exactly into a plano-concave lens (focal length f1, refractive index μ2, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be :

Question

A plano-convex lens (focal length f2, refractive index μ2, radius of curvature R) fits exactly into a plano-concave lens (focal length f1, refractive index μ2, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be : (A) f1 - f2 (B) f1 + f2 (C) (R/μ2 - μ1) (D) (2f1 f2/f1 + f2)

Tardigrade Admin · Accepted Answer

(1/F) = (1/f1) + (1/f2) = (1-μ1/R) + (μ2 - 1/R)

$f_{1} - f_{2}$

$f_{1} + f_{2}$

$\frac{R}{μ _{2} - μ _{1}}$

$\frac{2 f _{1} f _{2}}{f _{1} + f _{2}}$

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

f1​−f2​

f1​+f2​

μ2​−μ1​R​

f1​+f2​2f1​f2​​

F1​=f1​1​+f2​1​=R1−μ1​​+Rμ2​−1​

$f_{1} - f_{2}$

$f_{1} + f_{2}$

$\frac{R}{μ _{2} - μ _{1}}$

$\frac{2 f _{1} f _{2}}{f _{1} + f _{2}}$

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