Question

A 3.0-cm-tall object is placed 45.0 cm from a diverging lens having a focal length of magnitude 20.0 cm.

a) What is the distance between the image and the lens?

b) Is the image real or virtual?

c) What is the height of the image?

Answer 1

Given the height of the object (ho) = 3cm, the focal length of the diverging lens (f) = 20cm and the object distance (u) = 45cm

(a) By sign convention, f = -20cm and u = -45cm. If v is the image distance from the lens, then by lens equation

So the distance between the image and the lens is 13.85cm. Since the value is negative, the image is formed on the same side of the object.

(b) The magnification of the image is,

So the magnification of the image is 0.31. Since the magnification is positive, the image is virtual.

(c) If hi is the height of the image, then the magnification is given by,

So the height of the image is 0.93cm.