Class notes 12th grade NCERT Solutions - Biology for Class 12th

RAY OPTICS – ALL DERIVATIONS
Derivation of mirror formula

Consider an object AB placed on the principle axis beyond the centre of curvature C
of a concave mirror of small aperture, as shown.




Now A 'B 'C  ABC
A 'B' CB ' CP  B'P R  v
    .........(i)
AB BC BP  CP u  R
As A 'B 'C  APB, therefore,
A 'B 'P  ABP
Consequently,
A 'B' B 'P v v
   .................(ii)
AB BP u u
From (i) and (ii),we get
R  v v

u  R u
  uR  uv  uv  vR
 vR  uR  2uv
vR uR 2uv
  
uvR uvR uvR
1 1 2
  
u v R
As R  2f
1 1 2
so  
u v 2f
1 1 1
  
u v f

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Relation between focal length and radius of curvature for spherical mirror




From the diagram,

As AB is parallel to PC,

α  i
In BFC, r  α
Hence CF  FB
For a mirror small aperture
FB  FP
 CF  FP
hence
CP  CF  FP  FP  FP  2FP
or R  2f

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Derivation of thin lens formula