Question

A virtual image of magnification +5.0 is produced when an object is placed 5.0cm in front of a concave mirror. Find the radius of curvature of the mirror. Draw a ray diagram to verify your numbers.

Answer 1

given , magnification = 5

we have equation for magnification m = - di/do

where di is the distance of image

do is the distance of object

also given, do = -5 cm ( for a concave mirror , distance of object is taken negative)

substituting in m we get, 5= - di/ -5

5× -5 = -di

di = 25 cm

Now we have lens formula 1/f = 1/ do+ 1/ di

where f = focal length

substituting for do and di we get,

1/f = 1/-5 + 1/25

1/f = 1/25 -1/5

=( 5-25)/25 = -4/25

or f = -25/4 = -6.25 m

Our objective is to find radius of curvature.

we have , focal length , f = R /2

so R = 2f = -2×6.25 =- 12.5 m.

so radius of curvature is 12.5 m.

Ra rn obje 2 6.25 CHO