Question

Derive mirror formula for a convex mirror.

Answer 1

Case of Convex Mirror

Consider an object AB placed on the principal axis of a convex mirror. A ray AD is incident on the mirror at point D and is reflected along DX. It appears to come from F. Another ray AE is directed towards C and it retraces its path.

These rays of light appear to meet at A₁. Thus A₁B₁ is virtual, erect image of AB.

Draw DG ⊥ on principal axis.

∆s DGF and A₁B₁F are similar.

∴ \(\frac{DG}{A_1B_1} = \frac{GF}{B_1F}\)

But DG = AB

\(\frac{AB}{A_1B_1} = \frac{GF}{B_1F}\) ..............(i)

Again ∆s ABC and A₁B₁C are similar

∴ \(\frac{AB}{A_1B_1} = \frac{BC}{B_1C}\) ...........(ii)

Comparing (i) and (ii), we get

\(\frac{GF}{B_1F} = \frac{BC}{B_1C}\)

Since aperture of mirror is small, therefore, points G and D are close to point P

GF = PF

\(\frac{PF}{B_1F} = \frac{BC}{B_1C}\)

B₁C = PC - PB₁
BC = PB + PC
B₁F = PF - PB₁

\(\frac{PF}{PF-PB_1} = \frac{PB +PC}{PC-PB_1}\) ..............(iii)

Using sign coventions, we get

PF = +f, PB = -u, PB₁ = v, PC = +2f

Substituting in (iii), we get

\(\frac{+f}{+f-v} = \frac{-u+2f}{+2f-v}\)

2f² - vf = -uf + uv + 2f² - 2fv

uf + vf = uv

Dividing both sides by uvf, we get

\(\frac{uf}{uvf} + \frac{vf}{uvf} = \frac{uv}{uvf}\)

\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)