A) \[100\mu m\]
B) \[~300\mu m\]
C) \[1\mu m\]
D) \[~30\mu m\]
Correct Answer: D
Solution :
[d] \[\sin \theta =\frac{0.25}{25}=\frac{1}{100}\] Resolving power\[=\frac{1.22\lambda }{2\mu \sin \theta }=30\mu m\].You need to login to perform this action.
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