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

Options

A. 15cm
B. 20cm
C. `-18.31cm`
D. 10cm

Answer

c.

Explanation

The image `I^(')` for first refraction (i.e., when the ray comes out of liquid ) is at a depth of `=(33.25)/(1.33)=25cm`
`[ :' "Apparent depth" =("Real depth")/(mu)]`
Now, reflection will occur at concave mirror. For this `i^(')` behaves as an object
`:. u=-(15+25)-40cm, f=f, v=?`
Using mirror formula
`(1)/(f)=(1)/(v)+(1)/(u)=(1)/(v)-(1)/(u)-(1)/(40)rArr v=(40f)/(40+f)` (i)
But `v =-[15+(25)/(1.33)]` (ii)
where `(25)/(1.33)` is the real depth of the image
Form Eqs. (i) and (ii)
`(40f)/(40+f)=-[15+(25)/(1.33)]=-33.79`
`rArr f=-18.31cm`