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Question: In the given figure. The radius of curvature of curved surface for both the plano-convex and plano- ...

In the given figure. The radius of curvature of curved surface for both the plano-convex and plano- concave lens is 10 cm and refractive index for both is 1.5. The location of the final image after all the refractions through lenses is :

A

15 cm

B

20 cm

C

25 cm

D

40 cm

Answer

20 cm

Explanation

Solution

: The focal length of the plano – convex lens is

1f=(1.5−1)(1+10−1∞)=120\frac { 1 } { \mathrm { f } } = ( 1.5 - 1 ) \left( \frac { 1 } { + 10 } - \frac { 1 } { \infty } \right) = \frac { 1 } { 20 }

Focal length of plano –concave lens is

Since parallel beams are incident on the lens, its image from plano – concave lens will be formed at + 20 cm from it (at the focus) and will act as an object for the plano – concave lens. Since the two lens are at a distance of 10 cm from each other, therefore, for the next lens u = + 10 cm.

∴v=ufu+f=10×2010−20=20 cm\therefore \mathrm { v } = \frac { \mathrm { uf } } { \mathrm { u } + \mathrm { f } } = \frac { 10 \times 20 } { 10 - 20 } = 20 \mathrm {~cm}