Derivation or Proof of Mirror and lens formula CBSE(X) physics

Mirror formula is the relationship between object distance (u), image distance (v) and focal length.

1/v + 1/u = 1/f

In D ABC  and  D A’B’C

0

D ABC ~ D A’B’C [AA similarity]

AB /A’B’ = AC/A’C ----(I)

Similarly,

In DABC  and D A’B’C

0

D ABC ~ DA’B’C [AA similarity]

AB /A’B’ = AC/A’C ----(1)

Similarly, In D FPE  ~  D A’B’F

EP /A’B’ = PF/A’F

AB /A’B’ = PF/A’F   [ AB=EP]  ----(II)

From (i) &(ii)

AC/A’C =  PF/A’F  

=>  A’C/AC = A’F/PF

=> (CP-A’P)/(AP- CP)   = (A’P – PF)/PF

Now, PF = - f ;  CP = 2PF = -2f ; AP = -u ; and A’P = -v

Put these value in above relation:

[(-2f) –(-v)] /(-u)-(-2f) = {(-v) –(-f) }/(-f)

=>  uv = fv +uf

=>  1/f = 1/u + 1/v

Let AB is an object placed between f1 and f2 of the convex lens. The image A1B1 is formed beyond 2F2 and is real and inverted.

OA = Object distance = u ; OA1 = Image distance = v ; OF₂ = Focal length = f

In   D OAB and D OA₁B₁ 

D OAB ~  D OA₁B₁
A₁B₁ /AB = OA₁/OA -------------------(i)

Similarly, D OCF₂ ~  D F₂A₁B₁

A₁B₁ /OC = F₂A₁/OF₂

But we know that OC = AB

=>   A₁B₁ /AB = F₂A₁/OF₂ -------------------(ii)

From equation (i) and (ii), we get

OA₁/OA  = F₂A₁/OF₂

OA₁/OA  = (OA₁ - OF₂)/OF₂

v/-u = (v-u)/f

vf = -u(v-f)

vf = -uv + uf

Dividing equation (3) throughout by uvf

1/v  - 1/u = 1/f

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