Question

The size of image of an object placed at a distance 20cm from a concave mirror is four times the size of image produced when the same object is placed at a distance 45cm from the mirror.What is the focal length of the mirror if real image is produced in both the cases ?

Answer 1

Dear student
Let u be the object distance
u = 20cm u' = 45cm
let v be the image distance
then for u' = 45 cm ,v = 4v
​using mirror formula

$$\begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\ \frac{1}{4v}+\frac{1}{u\text{'}}=\frac{1}{f} \end{gathered}$$
focal length will remain the same,equating both the above equations.
$$\begin{gathered} \frac{1}{20}+\frac{1}{v}=\frac{1}{45}+\frac{1}{4v} \\ \frac{1}{20}-\frac{1}{45}=\frac{1}{4v}-\frac{1}{v} \\ 0.05-0.02=\frac{-3}{4v} \\ 0.03\times \frac{4}{3}=\frac{-1}{v} \\ 0.04=-\frac{1}{v} \\ v=-25cm \end{gathered}$$

to find te focal length
$$\begin{gathered} \frac{1}{20}-\frac{1}{25}=\frac{1}{f} \\ 0.05-0.04=\frac{1}{f} \\ 0.01=\frac{1}{f} \\ f=100cm \end{gathered}$$

Regards