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A microscope has an objective of focal length 1.5 cm and an eye piece of 2.5 cm. If the distance between objective and eye piece is 25 cm, is the value of magnification produced for relaxed eye is 50 n. Find the value of $n$ approximated to nearest integer.

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answer is 3.

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Detailed Solution

Here, $f_{0} = 1.5 c m, f_{e} = 2.5 c m$

Distance between objective and eye piece $= υ_{0} + f_{e} = 25 c m$

$∴ υ_{0} = 25 - f_{e} = 25 - 2.5 = 22.5 c m$

Using $\frac{1}{υ_{0}} - \frac{1}{u_{0}} = \frac{1}{f_{0}}$

$\frac{1}{22.5} - \frac{1}{u_{0}} = \frac{1}{1.5}$

$- \frac{1}{u_{0}} = \frac{1}{1.5} = \frac{1}{22.5} = \frac{21}{1.5 \times 22.5}$

$| u_{0} | = \frac{1.5 \times 22.5}{21} = 1.6 c m$

For relaxed eye, magnification, $M = \frac{υ_{0}}{u_{0}} \times \frac{D}{f_{e}} = \frac{22.5}{1.6} \times \frac{25}{2.5} = 140.6$

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