# Mirror Formula for Concave Mirror

Let us consider a ray of light AB strikes on the surface of a concave mirror at point B and is reflected back and passes through the point C. Let DB be the normal of the ray. Since angle of incidence is equal to the angle of reflection, we can say that AB = BC. Let α, β and γ be the angle made by the incident ray normal and the reflected ray with the reflecting surface. Now, again a ray of light passes through the principle axis P and strikes at the pole and returns in the same path. Then image is formed at the point I . Also, BE be the perpendicular line drawn from point B to the principle axis. Now,

By using basic geometrical laws,
α + i = β………..(i)
β + i = γ ………. (ii)
On solving equation (i) and (ii),
β – α = γ –β
2 β = γ + β…………(iii)

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

Now, in triangle BAE,

$$\text{Tan}\alpha = \alpha =\frac{\text{BE }}{\text{AE}}$$

In triangle BDE,
$$\text{Tan} \beta = \beta = \frac{\text{BE }}{\text{DE}}$$

Similarly, in triangle BCE,
$$\text{Tan} \gamma = \gamma = \frac{\text{BE }}{\text{EC}}$$

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

$$\frac{\text{2BE}}{\text{DP}} = {\frac{\text{BE}}{\text{CP}}} = {\frac{\text{BE}}{\text{AP}}}$$

$$or, \frac{\text{2}}{\text{DP}} = {\frac{\text{1}}{\text{CP}}} + {\frac{\text{1}}{\text{AP}}}$$

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

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

Equation (iii) is the required mirror formula.

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