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0.1 mm दूरी में पृथक दो बिन्दु एक सूक्ष्मदर्शी से बस देखें जा सकते हैं। जब 6000 Å तरंगदैर्ध्य का प्रकाश काम में लिया जा रहा है। यदि 4800 Å का प्रकाश काम में लिया लाए तो विभेदन सीमा क्या झेगी?

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एक सूक्ष्मदर्शी के लिए विभेदन सीमा

S = \(\frac{1.22\lambda y}{D} \times \alpha\)

अत: S2 = S\(\frac{\lambda_1}{\lambda}\)

= 0.1 x  \(\frac{4800}{6000}\)

= 0.8 mm

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