A card sheet divided into squares each of size 1 mm^{2} 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.

#### Solution

**(a) **Area of each square, A = 1 mm^{2}

Object distance, u = −9 cm

Focal length of a converging lens, f = 9 cm

For image distance v, the lens formula can be written as:

`1/"f" = 1/"v" - 1/"u"`

`1/10 = 1/"v" + 1/9`

`1/"v" = -1/90`

∴ v = −90 cm

Magnification, `"m" = "v"/"u"`

= `(-90)/-9`

= 10

∴ Area of each square in the virtual image = (10)^{2}A

= 10^{2} × 1

= 100 mm^{2}

= 1 cm^{2}

**(b) **Magnifying power of the lens = `"d"/|"u"| = 25/9` = 2.8

**(c) **The magnification in (a) is not the same as the magnifying power in (b).

The magnification magnitude is `(|"v"/"u"|)` and the magnifying power is `("d"/|"u"|)`.

The two quantities will be equal when the image is formed at the near point (25 cm).