Question

Consider two convex lenses of focal lengths f₁&f₂ are made from same material, are separated by a suitable distance d. The focal length of the combination is given by 1/f = 1/f₁+1/f₂-d/f₁f₂, the value of d for which combination is free from chromatic aberration is given by

a) (f₁+f₂)/2 

b) (f₁-f₂)/2 

c) (f₁+f₂) 

d) none

Answer 1

Consider two convex lenses of focal lengths f₁&f₂ separated by a suitable distance d. The focal length of the combination is given by: 1/f₁+1/f₂-d/f₁f₂ = 1/f (1) differentiate above equation

 -df₁/f₁² –df₂/f₂² –d[1/f₁(-df₂/f₂²)+1/ f₂(df₁/f₁²)] = (-1/f₂ )df {df=0 for Achromatism} 

(df₁/f₁)1/f₁+(df₂/f₂)1/f₂ –d[(df₂/f2)1/f₁f₂ +(df₁/f₁)1/f₁f₂]=0 

(ω₁/f₁) + ( ω₂/f₂) – d[(ω₂/f₁f₂) + (ω₁/f₁f₂)] =0 

(ω₁f₂+ ω₂f₁)/ f₁f₂ = d(ω₂+ω₁)/ f₁f₂ 

which gives d = (ω₁f1+ ω₂f₂)/ (ω₁+ω₂) 

In case when lenses are of same material, ω₁ = ω₂ = ω Then d = (f₁+f₂)/2 

Thus two lenses of same nature can be free from chromatic aberration if they are placed at a separation d = (f₁+f₂)/2