Question

Find the position of the object which when placed in front of a concave mirror of
radius of curvature 40 cm produces a virtual image, which is twice the size of the
object.

Answer 1

Answer:

Explanation:

given:

focal length

f= -40cm

and magnification

m = +2 = -v/u

or

v= -2u_______(1)

now, the mirror relation is

1/f = 1/v + 1/u

or

-1/40 = (u + v)/uv

= u-2u /(-2u)2 [substituting value of v from eq. (1)]

thus, image distance will be

u= -20cm

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Answer 2

Given, radius of curvature:- 40 cm

Thus, focal length= -20 cm

Also, magnification= -v/u= 2

v= -2u

mirror formula:- 1/v + 1/u = 1/f

Thus, f = vu/ v+ u

f= -2u×u/ -2u + u

f= -2u²/ -u

f= 2u= - 20

u = -10 cm

Thus position of object is at 10 cm from the mirror.

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