A concave mirror of focal length 10 cm forms an image which is double the size of object when the object is placed at two different positions. The distance between the two positions of the object is __________ cm.

Solution :

For mirrors,

$$m = -\frac{v}{u}$$

Also using mirror formula :

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

Combining both gives :

$$m = \frac{f}{f-u}$$

Given :

$$f = -10\text{ cm}$$

Image is double the size of object.

So two cases are possible :

Case 1 : Real inverted image

$$m = -2$$

Therefore,

$$-2 = \frac{-10}{-10-u}$$

$$20 + 2u = -10$$

$$2u = -30$$

$$u_1 = -15\text{ cm}$$

Case 2 : Virtual erect image

$$m = +2$$

Therefore,

$$2 = \frac{-10}{-10-u}$$

$$-20 - 2u = -10$$

$$-2u = 10$$

$$u_2 = -5\text{ cm}$$

Distance between the two object positions :

$$|u_1-u_2|$$

$$= |-15-(-5)|$$

$$= 10\text{ cm}$$

Final Answer :

$$10\text{ cm}$$