Question

An object of size 3.0cm is placed 14cm in front of a concave lens of focal length 21cm. Describe the image produced by the lens. Whathappens if the object is moved further away from the lens?

Answer 1

Given: The focal length of the concave lens is −21 cm, the object distance from lens is −14 cm and the size of the object is 3 cm.

The lens formula is given as,

1 f = 1 v − 1 u

where, the focal length of the lens is f, the distance of the object from the lens is u , and the distance of the image from the lens is v.

By substituting the given values in the above expression, we get,

1 −21 = 1 v − 1 −14 1 v =− 1 21 − 1 14 1 v = −2−3 42 v=−8.4 cm

Thus, the image is formed 8.4 cm away from the lens on the same side and negative sign shows that the image is virtual and erect.

Let m be the magnification of image, h 1 be the height of the object, and h 2 be the height of the image.

The magnification of the image is given as,

m= h 2 h 1

where, the magnification of image is m, the height of the object is h 1 and the height of the image is h 2 .

The magnification is also given as,

m= v u

From the above expressions,

h 2 h 1 = v u

By substituting the given values in the above expression, we get,

h 2 3 = −8.4 −14 h 2 =1.8 cm

Thus, the height of the image is 1.8 cm. If the object is moved further away from the lens, then virtual image will move towards the focus of the lens and the size of the object decreases.