Optics 7 Question 14
14. A container is filled with water $(\mu=1.33)$ up to a height of $33.25 cm$. A concave mirror is placed $15 cm$ above the water level and the image of an object placed at the bottom is formed $25 cm$ below the water level. The focal length of the mirror is
(2005, 2M)
(a) $10 cm$
(b) $15 cm$
(c) $20 cm$
(d) $25 cm$
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Solution:
- Distance of object from mirror
$$ =15+\frac{33.25}{1.33}=40 cm $$
Distance of image from mirror $=15+\frac{25}{1.33}$
$$ =33.8 cm $$
For the mirror, $\quad \frac{1}{v}+\frac{1}{u}=\frac{1}{f}$
$$ \begin{aligned} & \therefore & \frac{1}{-33.8}+\frac{1}{-40} & =\frac{1}{f} \\ & \therefore & f & =-18.3 cm \end{aligned} $$
$\therefore$ Most suitable answer is (c).
