Question

A wire mesh consisting of very small squares is viewed at a distance of $8 \,cm$ through a magnifying converging lens of focal length $10\, cm$, kept close to the eye. The magnification produced by the lens is

• A. 5
• B. 8
• C. 10
• D. 20

Answer 1

Answer: (A)

5

Answer 2

Lens formula is given by $\frac{1}{ f }=\frac{1}{v}-\frac{1}{ u }$ where $f$ is focal length of lens, $v$ is image distance and $u$ is object distance. Given, $f =10\, cm$ (as lens is converging) $u =-8 \,cm$ (as object is placed on left side of the lens) Substituting these values in E (i), we get $\frac{1}{10}=\frac{1}{ v }-\frac{1}{-8}$ $\Rightarrow \frac{1}{ v }=\frac{1}{10}-\frac{1}{8}$ $\Rightarrow \frac{1}{ v }=\frac{8-10}{80}$ $\therefore v =\frac{80}{-2}=-40\, cm$ Hence, magnification produced by the lens $m =\frac{ v }{ u }=\frac{-40}{-8}=5$