A container is filled with water 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 [

Question

A container is filled with water $(μ = 1.33)$ 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]

Smart Achievers · Accepted Answer

(3)

The image $I'$ for first refraction (i.e., when the ray comes out of the liquid) is at a depth of

$= \frac{33.25}{1.33} = 25 cm$ $[∵ Apparent depth = \frac{Real depth}{μ}]$

Now, reflection will occur at the concave mirror. For this, $I'$ behaves as an object.

Distance of object from mirror, $u = - (15 + 25) = - 40 cm$

and $v = - (15 + \frac{25}{1.33})$

where $\frac{25}{1.33}$ is the real depth of the image.

Using mirror formula,

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ $\Rightarrow \frac{1}{f} = \frac{1}{- 33.8} + \frac{1}{- 40}$

$∴$ $f = - 18.31 cm$

Solution

Q.
A container is filled with water $(μ = 1.33)$ 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]

1 15 cm

2 20 cm

3 −18.31 cm

4 10 cm

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