17 October 2013

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  A’B’C
0
ABC A’B’C [AA similarity]
AB /A’B’ = AC/A’C ----(I)
Similarly,
In DABC  and A’B’C
0
ABC DA’B’C [AA similarity]
AB /A’B’ = AC/A’C ----(1)
Similarly, In FPE  ~  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   OAB and OA1B1 
OAB ~  OA1B1
A1B1 /AB = OA1/OA -------------------(i)
Similarly, OCF2 ~  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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