(a) Mirror Formula: A formula which gives the relation between the image distance (v), the object distance (u) and the focal length (f) of a mirror is known as mirror formula
\(\frac{1}{\text {Focal length}} = \frac{1}{\text {Image distance}} + \frac{1}{\text {Object distance}} \)
\(\frac{1}{f} = \frac{1}{v} + \frac{1}{u}\)
(b) Assumptions
- Aperture of mirror is small.
- Incident ray makes small angles with principal axis.
- Object lies on the principal axis.
- Object lies on left hand side of the mirror.
Case of Concave Mirror
Consider an object AB placed beyond centre of curvature of a concave mirror, on its principal axis and on the left hand side.
A ray AD parallel to principal axis is incident on the mirror at point D and is reflected to pass through F. Another ray AE passing through centre of curvature C is reflected along the same path. Two rays of light intersect at point A1. Thus A1B1 is real, inverted and diminished image of AB formed between C and F.
Draw DG ⊥ on principal axis.
∆s DGF and A1B1F are similar,
∴ \(\frac{DG}{A_1B_1} = \frac{GF}{FB_1}\)
But DG = AB
\(\frac{AB}{A_1B_1} = \frac{GF}{FB_1}\) .............(i)
Again ∆s ABC and A1B1C are similar
\(\frac{AB}{A_1B_1} = \frac{CB}{CB_1}\)............(ii)
Comparing (i) and (ii)
\(\frac{GB}{FB_1} = \frac{CB}{CB_1}\)
Since the aperture is small, therefore, point D and point G lie
very close to P.
∵ GF = PF
\(\frac{PF}{FB_1} = \frac{CB}{CB_1}\) ...............(iii)
Since FB1 = PB1 - PF
CB = PB - PC
CB1 = PC - PB1
Substituting in (iii), we get
\(\frac{PF}{PB_1 - PF} = \frac{PB - PC}{PC - PB_1}\)
Applying sign conventions, we get
PF = -f, PC = -2f, PB = -u and PB1 = -v
\(\frac{-f}{-v(-f)} = \frac{-u-(-2)}{-2f-(-v)}\)
\(\frac{-f}{-v+f} = \frac{-u + 2f}{-2f+ v}\)
2f2 -vf = uv - uf - 2fv + 2f2
or uv - uf - 2fv = -vf
-uf - vf = -uv
uf + vf = uv
Dividing both sides by uvf, we get
\(\frac{uf}{uvf} + \frac{vf}{uvf} = \frac{uv}{uvf}\)
\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
