Question:
For the same objective, find the ratio of the least separation between two points to be distinguished by a microscope for light of 5000 Ao and electrons accelerated through 100V used as the illuminating substance.
Solution:
5000 Ao = 5000 × 10-10 m
1/d = 2 sin β/1.22λ
dmin = 1.22 λ/ 2 sin β
λd = 1.22/10 × 10-10 m
When the 100V light is used, d’min = 1.22 λd/ 2 sin β
d’min = 1.22 × 1.22 × 10-10/2 sin β
The required ratio = dmin/d’min = 1.22/5000 = 0.244 × 10-3
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