Question

A container is filled with water $(\mu =1.33)$ upto a height of $33.25cm$. A convex mirror is placed $15cm$ above the water level and the image of an object placed at the bottom is formed $25cm$ below the water level. Focal length of the mirror is

• A. $15cm$
• B. $20cm$
• C. $-18.31cm$
• D. $10cm$

Answer 1

The correct option is C $-18.31cm$

The image $I^{\prime }$ for first refraction (i.e., when the ray comes out of liquid) is at a depth of $=\frac{33.25}{1.33}=25cm$
$[\because Apparentdepth=\frac{Realdepth}{\mu }]$
Now, reflection will occur at concave mirror. For this $I^{\prime }$ behaves as an object
$\therefore u=-(15+25)=-40cm,f=f,v=?$
Using mirror formula
$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{v}-\frac{1}{40}\Rightarrow v=\frac{40f}{40+f}...\left(i\right)$
But $v=-[15+\frac{25}{1.33}]...\left(ii\right)$
where $\frac{25}{1.33}$ is the real depth of the image.
From Eqs. (i) and (ii),
$\frac{40f}{40+f}=-[15+\frac{25}{1.33}]=-33.79$
$\Rightarrow f=-18.31cm$.