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एक संयुक्त सूक्ष्मदर्शी के अभिदृश्यक की फोकस-दूरी 1 cm और नेत्रिका की फोकस-दूरी 2 cm है | यदि नली की लम्बाई 20 cm हो, तो यंत्र की आवर्धन-क्षमता निकाले |

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संयुक्त सूक्ष्मदर्शी की आवर्धन-क्षमता,
`M = ((L)/(f_(0)))((D)/(f_(e))),` जहाँ D स्पष्ट दर्शन की न्यूनतम दूरी = 25 cm है |
यहाँ, `L = 20 cm, f_(0) = 1 cm` तथा `f_(e) = 2 cm`.
`therefore M = ((20)/(1))xx((25)/(2))=250.`
अतः, यंत्र की आवर्धन-क्षमता ` = 250.`

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