Question

A converging lens of focal $20$ cm and diameter $5$ cm is cut along the line $AB$. The part of the lens shown shaded in the diagram is now used to form an image of a point $P$ placed $30$ cm away from it on the line $XY$. Which is perpendicular to the plane of the lens. The image of P will be formed.

• A. $0.5$ cm above $XY$
• B. $1$ cm below $XY$
• C. on $XY$
• D. $1.5$ cm below $XY$

Answer 1

The correct option is D $1.5$ cm below $XY$

Let an object is placed on the principal axis at a distance of $30cm$ from the lens such that its top most point is at point $P$ . As point $P$ is $0.5cm$ above the principal axis , therefore height of the object will be ,

$O=0.5cm$ ,

given - $f=20cm,u=-30cm$

by lens formula ,

$v=\frac{uf}{u+f}$

$v=\frac{-30\times 20}{-30+20}=60cm$

now we have , $I/O=v/u$ ,

$I/0.5=60/-30$

$I=-1cm$

Negative sign shows that image will be below the principal axis (inverted). $1cm$ is the distance of top most point of image from principal axis and distance between XY and principal axis is $0.5cm$ therefore distance of top most point of image (image of point $P$) from XY will be $=1+0.5=1.5cm$ (below XY).