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 10 cm) held close to the eye. What is the magnification produced by the lens? How much is the area of each square in the virtual image?

Solution

Given,
Object distance, u = –9 cm
Focal length of converging lens, f = 10 cm

Now, Using the formula,

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

\Rightarrow

\frac{1}{v} = \frac{1}{f} + \frac{1}{u}

\Rightarrow

\frac{1}{v} = \frac{1}{10} + \frac{1}{- 9} = \frac{- 9 + 10}{- 90} = - \frac{1}{90}

i.e.,

v = - 90 cm

Magnification, m

= \frac{v}{u} = \frac{- 90}{- 9} = 10

Therefore,

Area of the each square in the virtual image=(10 x 10 x 1) mm²= 100 mm² = 1 cm²

For Daily Free Study Material Join wiredfaculty