An object is placed in front of a concave mirror at a distance of 7.5 cm from it, image is formed at a distance of 30 cm from the mirror. Find the focal length of mirror when (i) Real image is formed (ii) Virtual image is formed.

Solution: For a concave mirror u becomes negative, $$\therefore u=-7.5cm$$ (i) When real image is formed v is also negative $$\therefore v=-30cm$$

By using mirror formula, $$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$$ Lets substitute the values, $$\frac{1}{-7.5}+\frac{1}{-30}=\frac{1}{f}$$ $$\therefore \frac{1}{f}=\frac{-30-7.5}{(-7.5)\times (-30)}=\frac{-37.5}{225}$$ $$\therefore f=\frac{225}{-37.5}=-6cm$$ (ii) When virtual image is formed v is positive $$\therefore v=30cm$$ By using mirror formula, $$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$$ Lets substitute the values, $$\frac{1}{-7.5}+\frac{1}{30}=\frac{1}{f}$$ $$\therefore \frac{1}{f}=\frac{30-7.5}{(-7.5)\times \left(30\right)}$$ $$=\frac{22.5}{-225}$$ $$\therefore f=\frac{-225}{22.5}=-10cm$$ Hence the focal length of mirror when real image is formed is -6 cm and when virtual image is formed at 30 cm, focal length of mirror will be -10 cm.