Question

An object is 23 cm in front of a diverging lens that has a focal length of -11 cm. How far in front of the lens should the object be placed so that the size of its image is reduced by a factor of 1.9?

Answer 1

Use thin lens equation for diverging lens that has a focal length of -11 cm and object distance u=-23 cm. Our goal is to find the image distance v:

$$\frac{1}{v}-\frac{1}{u}=\frac{1}{f},\\\space\\ v=-\bigg(\frac{1}{12}+\frac{1}{23}\bigg)^{-1}=-7.44\text{ cm}$$

Magnification is

$$m=\frac{v}{u}=0.324.$$

This is how is reduced initially. Finally, it must be 1.9 times smaller:

$$m'=\frac{v'}{u'}=\frac{m}{1.9}=0.170.$$

What is the new object distance?

$$\frac{1}{v'}-\frac{1}{u'}=\frac{1}{f},\\\space\\ \frac{1}{m'u'}-\frac{1}{u'}=\frac{1}{f},\\\space\\ u'=\frac{f(1-m')}{m'}=\frac{-11(1-0.17)}{0.17}=-53.7\text{ cm}.$$