1/v + 1/u = 1/f

In Triangle ABC and Triangle A’B’C

<A = <A’ = 900

<C =<C ( vert. opp. <s]

Triangle ABC ~Triangle A’B’C [AA similarity] => AB /A’B’ = AC/A’C ----(I)

Similarly,

In Triangle ABC and A’B’C

<A = <A’ = 900

<C =<C ( vert. opp. <s]

Also, in Triangle ABC ~Triangle A’B’C [AA similarity]

AB /A’B’ = AC/A’C ----(1)

Similarly, In DFPE ~ 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

you may also see (a) Mirror formula (b) Mirror Formula

Derivation or Proof-of-Mirror and Lens formula(X) physics |

**Derivation or Proof-of- the relation between focal length and radius of curvature(X) physics [R=2f]**

Consider a ray of light AB, parallel to the principal axis, incident on a spherical mirror at point B. The normal to the surface at point B is CB and CP = CB = R, is the radius of curvature. The ray AB, after reflection from mirror will pass through F (concave mirror) or will appear to diverge from F (convex mirror)

From the figure,

According to law of reflection,< i = <r

<i = <q [Since, AB IICP]

=> <r= <q

So, In D BCF, CF = BF ----(i)

If the aperture of the mirror is small, B lies close to P, BF = PF (ii)

From (i) and (ii)

CF = FP

Now, PC = PF + FC = 2 PF

or R = 2f