The semi vertical angle subtended by two points of an object at the objective of a microscope is `30^@`. The object is illuminated using light of wavelength 589.3 nm. Calculate the minimum distance between the two points if these are just resolved. If a medium of `R.I 1.48` is used between the object and the objective, then calculate the new limit of resolution. Comment on the improvement in the limit of resolution.

Explanation

Given `theta=30^@, lambda=5.89xx10^(-7) m` .
We know that `dx=(1)/(2n sin theta)`
i.e., `(dx)_("air")=(5.893xx10^(-7))/(2xx1xxsin 30^@)`
`(du)_("air")=5.893xx10^(-7) m `
If a transparent medium is used then
`(dx)_("med")=((dx)_("air))/(n)=(5.893xx10^(-7))/(1.48)m=3.982xx10^(-7) m .`
`Delta` limit of resolution `-(5.893-3.982)xx10^(-7)m=1.911 xx10^(-7) m. `
Percentage improvement `=(1.911)/(5.893)xx100=32.43%`
i.e., the instrument can resolve even if the distance between the points is decreased by `32.43 %`.