Question

A container is filled with water (n = 1.33) up to a height of 33.25 cm. A concave mirror is held 15 cm above the water level, and the image I of an object O placed at the bottom is formed 25 cm below the water level. The focal length of the mirror is roughly:

पर्याय

• 10 cm

• 15 cm

• 20 cm

• 25 cm

Answer 1

20 cm

Explanation:

Given:

Refractive index of water (n) = 1.33

Depth of water = 33.25 cm

Mirror is 15 cm above the water.

Image is formed 25 cm below the water surface.

Object is placed at the bottom of the tank.

A detailed step-by-step breakdown is given below:

1. Real depth of object O:

The object is at the bottom, so real depth = 33.25 cm

2. Apparent depth due to refraction:

Apparent depth = `"Real depth"/n`

= `33.25/1.33`

≈ 25 cm

So, due to refraction, the object appears to be 25 cm below the water surface to the mirror.

3. Distance of the apparent object from the mirror:

Since the mirror is 15 cm above the water surface, the object appears to be at a distance:

u = −(25 + 15) = −40 cm  ....(negative because the object is in front of the mirror)

4. Image is formed 25 cm below the water surface:

That is, the image is at 25 cm from the water surface and 15 cm + 25 cm = 40 cm from the mirror.

So, v = −40 cm

5. Using the mirror formula:

`1/f = 1/v + 1/u`

= `1/(-40) + 1/(-40)`

= `(-2)/40`

= `-1/20`

f = −20 cm

Thus, the focal length of the mirror is approximately 20 cm.