The focal lengths of objective and eye-lens of a microscope are 1cm and 5cm respectively magnifying power for the relaxed eye is 45 then the length of the tube is:
30cm
15cm
12cm
25cm
A
25cm
B
15cm
C
12cm
D
30cm
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Solution
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fobjective =f0=1cm feye peice =fe=5cmm2=45m= magnifieng power for relaxed eye Δ= least D istance of vision for normal eye ie 25cmL∞= Length of tube for m∞.
By using the formula: m∞=(L∞−f0−fc)×Df0×fe45=(L∞−1−5)×251×59=L∞−6L∞=15cm
Hence option (c) is correct
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