Question

A 4 cm tall object is placed perpendicular to the principal
20 cm. The distance of the object from the lens is 15 cm
placed perpendicular to the principal axis of a convex lens of focal length image
​

Answer 1

Answer:

Given - f=+24cm , u=-16cmf=+24cm , u=−16cm ,

As |u|(16cm)<|f|(24cm)∣u∣(16cm)<∣f∣(24cm) ,it means object lies between FF and CC , in this position of object the rays from object cannot meet at any point on the other side of lens ,which is also clear from the position of image found below ,

From lens equation v=\dfrac{uf}{u+f}=\dfrac{-16\times24}{-16+24}=-48cmv=

u+f

uf

=

−16+24

−16×24

=−48cm ,

-ive sign shows that image will be formed on the same side of lens , where the object is placed .

Now linear magnification m=I/O=v/u=-48/-16=+3m=I/O=v/u=−48/−16=+3 ,

or I=3\times4=12cmI=3×4=12cm (given O=4cmO=4cm) ,

since m=+ivem=+ive and m>1m>1 ,therefore image will be virtual ,erect and magnified .