A plano convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lens are made of different materials of refractive indices $μ_{1}$ and $μ_{2}$ $R$ is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is
$\frac{2 R}{M_{2} - M_{1}}$
$\frac{R}{2 (M_{2} + M_{2})}$
$\frac{R}{2 (M_{1} - M_{2})}$
$\frac{R}{{(M_{1} - M_{2})}_{1}}$

Question

A plano convex lens fits exactly into a plano-concave lens. Their plane surfaces are parallel to each other. If lens are made of different materials of refractive indices $μ_{1}$ and $μ_{2}$ $R$ is the radius of curvature of the curved surface of the lenses, then the focal length of the combination is
$\frac{2 R}{M_{2} - M_{1}}$
$\frac{R}{2 (M_{2} + M_{2})}$
$\frac{R}{2 (M_{1} - M_{2})}$
$\frac{R}{{(M_{1} - M_{2})}_{1}}$

Toppr Admin · Accepted Answer

1F-x3BC-1-x2212-1-x2212-1R-1F-x2212-x3BC-2-x2212-1-1R-1re-x2212-x3BC-1R-1R-x3BC-2R-x2212-1R1FC-x3BC-2-x2212-x3BC-1RF-R-x3BC-2-x2212-x3BC-1-xA0-Hence option -D- is correct

1F=(μ1,−1)(−1R)1F−(μ2−1)(1R)1re=−μ1R+1R+μ2R−1R1FC=μ2−μ1RF=Rμ2−μ1 Hence option (D) is correct