Question
If the red light is replaced by blue light illuminating the object in a microscope the resolving power of the microscope
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Answer
(b) $R.P.$ $ \propto \frac{1}{\lambda } \Rightarrow \frac{{{{(R.P.)}_1}}}{{{{(R.P.)}_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{5}{4}$
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