Question

Find the position of an object which when placed in front of a concave mirror of focal length 10cm produces a virtual image which is twice the size of the object.​

Answer 1

Position of object (u) = -5 cm

Explanation:

Given,

Magnification ( m) = 2 (virtual)

Focal length (f) = - 10 cm

Image distance = + v (virtual)

Object distance = u

We know that,

m = -v/u

=> 2 = -v/ u

=> v = -2u ...(1)

ALSO,

1/f = 1/v + 1/u

=> -1/10 = -1/2u + 1/u .{from(1)}

=> -1/10 = ( -1 + 2)/2u

=> -1/10 = 1/2u

By cross multiplication we get,

=> -2u = 10

=> u = -5

Hence, position of object = - 5 cm

Answer 2

Answer:

m=-v/u =2

Explanation:

Object distance with sign is -22.5 cm. Negative sign indicates the object is in front of the mirror. Hence objects must be placed 20cm in front of the mirror to get the magnification 3. Hence option A will be the answer.