Question

A CHALLENGING QUESTION FOR CLASS 10 BRAINLIEST USERS;
An object of size 5cm is placed at a distance of 20cm in front of a convex lens of focal length 15cm. At what distance from the lens should a screen be placed so that a sharp image can be obtained.
Find the size and nature of the image obtained.
Give reasons for the signs taken either positive or negative..

It’s urgent guys!
Help me out!

Answer 1

HOPE IT HELPS

MARK AS BRAINLIEST ❤

$\infty$

Answer 2

object distance =- 20cm

object size= 5cm

focal length = 15cm

image distance=?

image size =?

A/Q

lens formula =

$\frac{1}{f } = \frac{1}{v} - \frac{1}{u} \\ \frac{1}{v} = \frac{1}{f} + \frac{1}{u}$

The image distance is 60cm & size is -15cm.

The +ve sign of image distance or negative sign of magnification shows that real,inverted & diminished image is formed.