Question

A dime is placed in front of a concave mirror that has a radius of curvature R = 0.40 m. The image of the dime is inverted and two times the size of the dime. Determine the distance between the dime and the mirror.

Answer 1

distance between the dime and the mirror, u = 0.30 m

Given:

Radius of curvature, r = 0.40 m

magnification, m = - 2 (since,inverted image)

Solution:

Focal length is half the radius of curvature, f =
`(r)/(2)`

f =
`(0.40)/(2) = 0.20 m`

Now,

m = -
`(v)/(u)`

- 2 = -
`(v)/(u)`

`(v)/(u)` = 2 (2)

Now, by lens maker formula:

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

`(1)/(v) = (1)/(f) - (1)/(u)`

v =
`(uf)/(u - f)` (3)

From eqn (2):

v = 2u

put v = 2u in eqn (3):

2u =
`(uf)/(u - f)`

2 =
`(f)/(u - f)`

2(u - 0.20) = 0.20

u = 0.30 m