# Mirror Formula for Convex Mirror

Consider a ray of light AB incident on a convex at point B and is then reflected back through BC. The ray of light is seems to be coming from IB. let us draw a perpendicular distance from point B to principle axis. While another ray of light passes through the principle axix and is then reflected back in its own path. Image I is formed by the intersection of two rays of light. Now, by using basic geometrical rules:

∠DBA = ∠BAF + ∠BFA

θ = α + β……… (i)

Also, ∠BIG = ∠IBF + ∠ BFI

γ = θ + β……… (ii)

Now, from equation (i) and (ii),

γ – β = α + β

2 β = γ – α……. (iii)

Since the value of α, β and γ is very small, we can write Tanα =α, Tanβ =β and Tanγ= γ.

Now, in triangle ABG,

$$ \text{Tan} \alpha = \alpha = \frac{\text{BG}}{\text{AG}}$$

In triangle BGF,

$$ \text{Tan} \beta = \beta = \frac{\text{BG }}{\text{GF}}$$

In triangle BGI,

$$\text{Tan} \gamma = \gamma = \frac{\text{BG }}{\text{GI}}$$

Since, the aperture of a mirror is very small, so the point G lies nearly to the point P,

So, we can write,

$$ \alpha = \frac{\text{BE }}{\text{AP}}$$

$$ \beta = \frac{\text{BE }}{\text{DP}}$$

$$ \gamma = \frac{\text{BE }}{\text{CP}}$$

Now, putting the value of α, β and γ in equation (iii) we get,

$$ \frac{\text{2BG}}{\text{PF}} = \frac{\text{BG}}{\text{PI}} – {\frac{\text{BG}}{\text{AP}}}$$

Now, using sign convention,

$$ \frac{\text{-2}}{\text{PF}} = {\frac{\text{-1}}{\text{PI}}} + {\frac{\text{-1}}{\text{AP}}}$$

$$ or, \frac{\text{1}}{\text{u}} = {\frac{\text{1}}{\text{v}}} + {\frac{\text{2}}{\text{R}}}$$

$$ \therefore \frac{\text{1}}{\text{f}} = {\frac{\text{1}}{\text{u}}} + {\frac{\text{1}}{\text{v}}}……(iv)$$

Equation (iv) is the mirror formula for convex mirror.

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