Two identical convex lenses $L_{1}$ and $L_{2}$ are placed at a distance of $20 c m$ from each other on the common principal axis. The focal length of each lens is $15 c m$ and the lens $L_{2}$ is to the right of lens $A$ .A point object is placed at a distance of $20 c m$ on the left of lens $L_{1}$ , on the common axis of two lenses. Find, where a convex mirror of radius of curvature $5 c m$ should be placed to the right of $L_{2}$ so that the final image coincides with the object?

Question

Verified by Toppr

Answer 1

The distance $= 20 c m$

Focal lengths of each lense $- 15 c m$

The lens $l_{2}$ to the right of lens A

A point object is placed at a distance of $= 20 c m$ left of lose $l_{1}$

$r = 5 c m$

Image after 1^st refraction from lens

$\frac{1}{v} - \frac{1}{( - 2 0 )} = \frac{1}{1 5} v = 60 c m$

After reflection from the mirror,

$v^{'} = - 60 c m$

It will refract again by the curved surface

$\frac{1}{v ^{”}} - \frac{1}{6 0} = \frac{1}{1 5} \frac{1}{v ^{”}} = 12 c m 12 c m l e f t$

Answer 2

The distance $= 20 c m$

Focal lengths of each lense $- 15 c m$

The lens $l_{2}$ to the right of lens A

A point object is placed at a distance of $= 20 c m$ left of lose $l_{1}$

$r = 5 c m$

Image after 1^st refraction from lens

$\frac{1}{v} - \frac{1}{( - 2 0 )} = \frac{1}{1 5} v = 60 c m$

After reflection from the mirror,

$v^{'} = - 60 c m$

It will refract again by the curved surface

$\frac{1}{v ^{”}} - \frac{1}{6 0} = \frac{1}{1 5} \frac{1}{v ^{”}} = 12 c m 12 c m l e f t$