JEE Main Physics Class 12 Ray Optics and Optical Instruments Part 2 Questions

The given situation is shown in the following ray diagram. In concave mirror, when object is placed beyond C the image isformed between C and focus of mirror.
Consider radius of curvature of mirror is R.
Object distance, u=−(R+d1)
Image distance, v=−(R−d2)
Focal length of mirror, f=

−R

2

Now, from mirror formula, we have

1

f

=

1

v

+

1

u

=

u+v

uv

⇒uv=f(u+v)
Substituting the values in above expression, we get
[−(R+d1)][−(R−d2)] =−

R

2

(R+d1+R−d2)
⇒2(R2−d2R+d1R−d1d2) =2R2+d1R−d2R
⇒2R2−2d2R+2d1R−2R2−d1R+d2R =2d1d2
⇒d1R−d2R=2d1d2
⇒R=

2d1d2

d1−d2

Thus, radius of curvature of mirror is

2d1d2

d1−d2

.