A) \[0.24\mu m\]
B) \[0.48\mu m\]
C) \[0.12\mu m\]
D) \[0.38\mu m\]
Correct Answer: A
Solution :
Numerical aperature of the microscope is given as\[NA=\frac{0.61\lambda }{d}\] Where d = minimum sparation between two points to be seen as distinct \[d=\frac{0.61\lambda }{NA}=\frac{(0.61)\times (5000\times 10{{m}^{-10}})}{1.25}\] \[=2.4\times {{10}^{-7}}m\] \[=0.24\mu m\]
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