Question

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

Answer 1

(a) Area of each square, A=1mm²
Object distance, u= -9cm
The focal length of a converging lens, f= 10cm
For the image distance v, the lens formula can be written as: \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
\(\frac{1}{10}=\frac{1}{v}+\frac{1}{9}\)
\(\frac{1}{v}=-\frac{1}{90}\)
∴ v = -90cm
Magnification, m = \(\frac{v}{u}\) = \(\frac{-90}{-9}\) = 10
Area of each square in the virtual image = (10) 2A = 10² × 1 = 100mm² = 1cm²
(b) Magnifying power of the lens = \(\frac{d}{|u|}\)= \(\frac{25}{9}\)= 2.8
(c) The magnification in (a) is not the same as the magnifying power in (b). The magnification power is (|v/u|) and the magnifying power is \(\frac{d}{|u|}\)). The two quantities will be equal when the image is formed at the near point (25cm).