Light of wavelength 589 nm is used to view an object under a microscope. The aperature of the objective has a diameter of `0.900 cm`. Find
What effect would this have on the resolving power, if water `(mu = 1.33)` fills the space between the object and objective.

Explanation

Here, `lambda 589nm = 589xx 10^(-9)m`
`a = 0.900cm = 0.900xx 10^(-2)m`
(or) `triangle theta = 1.22(lambda/a)= 1.22((589xx10^(-9)m)/(0.900xx10^(-2)m)) rad`
`= 7.89xx 10^(-5)rad`