A card sheet divided into squares each of size 1 mm2 is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 9 cm) held close to the eye.
(a) What is the magnification produced by the lens? How much is the area of each square in the virtual image?
(b) What is the angular magnification (magnifying power) of the lens?
(c) Is the magnification in (a) equal to the magnifying power in (b)? Explain.
Given:
Area of each square, A = 1 mm2
Object distance, u = −9 cm
Focal length of a converging lens, f = 10 cm
(a) Applying the lens formula we have:
…(1)
Where, f0 = focal length of the objective lens
v0 = Distance of image formation
u0 = Distance of object from objective
The lens formula for the image distance v, can be written as:
…(2)
By putting available values in equation (2) we get,
⇒ V0 = -90 cm
Magnification, m = v/u
m = −90/−9
m = 10
Therefore, area of each square in the virtual image = (10)2 A
= 102 x 1 = 100 mm2
= 1 cm2
(b) The lens has a magnifying power of = d/ |u|
= 25/9
= 2.8
(c) The magnification in (a) is not the same as the magnifying power in (b). Both the quantities will be same once the image is formed at the near point (25 cm)