Question

. A 4.0cm tall object is placed perpendicular to the principal axis of a convex lens of
focal length 0.1m the distance of the object from the lens is 150mm.
Find the nature, position and height of the image formed.​

Answer 1

Answer:

image will be virtual ,erect and magnified .

Explanation:

Given  -    f=+24cm,u=−16cm ,

As ∣u∣(16cm)<∣f∣(24cm) ,it means object lies between F and C , 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=

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=+3 ,

or                                        I=3×4=12cm       (given  O=4cm) ,

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