1.

For the same objective , the ratio of the least separation between two points to be distinguisghed by a microscope for light of 5000 Å and electrons accelerated through 100 V used as the illuminating substance is

Answer»

`10^(-3)`
`0.5 xx 10^(-3)`
`0.2 xx 10^(-3)`
`10^(-4)`

Solution :`d_(min) =(1.22 lambda)/(2 sin beta) `
where `beta` is the angle subtended by the OBJECTIVE atthe object.
For light of `5500 Å`
`d_(min) = (1.22 xx 5.5 xx 10^(-7))/(2 sin beta) m`
For electrons accelerated through 100 V the de Brogile WAVELENGTH is
`lambda = (h)/(p) = (1.227)/(SQRT(100)) = 0.13 nm = 0.13 xx 10^(-9)`m
`THEREFORE d._(min) = (1.22 xx 1.3 xx 10^(-10))/(2 sin beta)`
`(d._(min))/(d_(min)) = (1.3 xx 10^(-10))/(5.5 xx 10^(-7)) approx 0.2 xx 10^(-3)`


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