Question

How far should be an object be placed from the pole of mirror of focal length 20 centimeter to form image of size exactly 1 by fourth size of the object​

Answer 1

Answer: 100 cm

Explanation:

we know that magnification m = v / u , where v is the image distance and u is object distance.

So,  1/4 = v / u ⇒ v = u /4

Using the mirror formula:

$\frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

$or,\frac{1}{u} + \frac{4}{u} = \frac{1}{f} \\\\or, \frac{5}{u} = \frac{1}{f} \\\\or, u = 5 \times f = 5 \times 20 cm = 100 cm$

So the object should be placed at a distance of 100 cm from the pole of mirror.

Answer 2

Answer:

   u = -100cm

Explanation:

focal length (f) = -20cm

Magnification (M) = 1 / 4

image distance (v) = v

object distance (u) = u

M = -v / u

Here magnification is negative because it forms a real image

This is because virtual image formed by concave mirror is enlarged whereas according to the question the image is diminished and hence it forms a real image

-(1 / 4) = -v / u

1 / 4 = v / u

u = 4v -------{i}

1 / v + 1 / u = 1 / f

1 / v + 1 / 4v = - 1 / 20

4 / 4v + 1 / 4v = -1 / 20

5 / 4v = -1 / 20

4v = -100

v = -25cm

u = 4v = -100cm

∴ The object should be placed 100 cm to the left of the mirror.