Question

An object is 39 cm front of a converging lens with a focal length of 28 cm. What is the magnification of the lens

Answer 1

Solution. Using lens formula

$$\frac{1}{u}+\frac{1}{v}=\frac{1}{F}$$

where u is object distance; v is image distance; F is focal length.

According to the condition of the problem u=0.39m, F=0.28m. Therefore

$$\frac{1}{v}=\frac{1}{F}-\frac{1}{u}$$ $$v=\frac{Fu}{u-F}=\frac{0.39\times0.28}{0.39-0.28}\approx0.993m$$

Using the figure, the magnification is

$$M=\frac{h'}{h}=\frac{v}{u}=\frac{0.993}{0.39}\approx2.54$$

Answer. M=2.54