For the same objective, the ratio of least separation between two points to be distiguised by a microscope for light of `5000 Å` and electrons acclerated through 100V used as illuminating substance is __________ (neraly)

Explanation

We known that
Resolving power `=(1)/(d)=(2 sin beta)/(1.22 lambda) implies d_("min")=(122 lambda)/(2 sin beta )`
where, `lambda` is the wavelenght of light and `beta` is the angle substended by the objective at the object.
For the light of wavelength 5500Å.
`d_("min")=(1.22xx5500xx10^(-10))/(2 sin beta )`…..(i)
For electrons accelerated through 100 V, the de-Broglie wavelength
`lambda=(12.27)/(sqrt(V))=(12.27)/(sqrt(100))=0.12xx10^(-9) m`
`d_("min")=(122xx0.12xx10^(-9))/(2 sin beta )`
Ratio of the least seperation
`:. (d'_("min"))/(d_("min"))=(0.12xx10^(-9))/(5500xx10^(-10))=0.2xx10^(-3)`