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Consider three rays from the tip B of the object AB incident on the convex mirror, one ray passing through the center of curvature C, the second through the optical center and the third parallel to the principal axis.

The reflected ray of the third incident ray passes through the focus F and makes triangle CB¹F with the first incident ray.

Let the distance of C and F from the pole P be R and f respectively.

An inverted image A¹B¹ is formed at the point of convergence of all the reflected rays and A¹ acts as a perpendicular to the principal axis.

u = object distance ; v = image distance.

ΔABC ≈ ΔA¹B¹C

A¹B¹/AB = A¹C/AC...............(1)

ΔABD ≈ ΔA¹B¹P

A¹B¹/AB = PA¹/PA...............(2)

From (1) and (2),

A¹C/AC = PA¹/PA

PC¹ - PA / PA - PC = PA¹/PA ⇒ -R + v / -u + R = -u/-v = u/v

-uR + uv = -uv + vR ⇒ 2uv = uR + vR

Divide with uvr.

2/R = 1/v + 1/u

But R= 2f so..2/R = 1/f

⇒

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Mirror
formula is the relationship between

object distance (u), image distance (v) and focal length (f).

A ray of light starting from A and incident on the mirror along AD parallel to the principal axis of the mirror and gets reflected from the mirror and passes through F.

Another ray of light incident along AP is reflected along PA’ such that angle APB= angle i= angle BPA’= angle r.

The two reflected ray meet at A’ which is a real image.

Therefore A’B’ is real, inverted image of AB formed by reflection from the mirror.

In Triangles BAC and A’B’C’

AB/A’B’ = CB/CB’ ……………. (i)

Now Triangle ABP and A’B’P’ are similar,

AB/A’B’ = PB/PB’ …………………. (ii)

From (i) & (ii) … CB/CB’= PB/PB’ ……………….. (iii)

Measuring all distances from P,

we have CB= PB-PC CB’= PC-PB’

Therefore from (iii),

(PB-PC)/(PC-PB’) = PB/PB’ …………… (iv)

Using new cartesian sign convections,

PB= -u (distance of object)

PC= -R PB’ = -v (distance of image)

We get from (iv),

(-u+R)/(-R+v) = -u/-v

Or, +uR – uv = uv – vR

Or, uR + vR = 2uv

Dividing both sides by uvR,

We get, 1/v + 1/u = 2/R

As R=2f

Therefore, 1/v + 1/u = 2/R = 2/2f = 1/f This is the required formula.

object distance (u), image distance (v) and focal length (f).

A ray of light starting from A and incident on the mirror along AD parallel to the principal axis of the mirror and gets reflected from the mirror and passes through F.

Another ray of light incident along AP is reflected along PA’ such that angle APB= angle i= angle BPA’= angle r.

The two reflected ray meet at A’ which is a real image.

Therefore A’B’ is real, inverted image of AB formed by reflection from the mirror.

In Triangles BAC and A’B’C’

AB/A’B’ = CB/CB’ ……………. (i)

Now Triangle ABP and A’B’P’ are similar,

AB/A’B’ = PB/PB’ …………………. (ii)

From (i) & (ii) … CB/CB’= PB/PB’ ……………….. (iii)

Measuring all distances from P,

we have CB= PB-PC CB’= PC-PB’

Therefore from (iii),

(PB-PC)/(PC-PB’) = PB/PB’ …………… (iv)

Using new cartesian sign convections,

PB= -u (distance of object)

PC= -R PB’ = -v (distance of image)

We get from (iv),

(-u+R)/(-R+v) = -u/-v

Or, +uR – uv = uv – vR

Or, uR + vR = 2uv

Dividing both sides by uvR,

We get, 1/v + 1/u = 2/R

As R=2f

Therefore, 1/v + 1/u = 2/R = 2/2f = 1/f This is the required formula.