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Q.

The focal length of objective and eye lens of a microscope are $4 c m$ and 8 cm respectively. If the least distance of distinct vision is $24 c m$ and object distance is $4.5 c m$ from the objective lens, then the magnifying power of the microscope will be

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a

64

b

32

c

20

d

18

answer is B.

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Detailed Solution

$\frac{1}{f_{0}} = \frac{1}{v_{0}} - \frac{1}{u_{0}}$

$\Rightarrow \frac{1}{(+ 4)} = \frac{1}{v_{0}} - \frac{1}{(- 4 .5)} \Rightarrow v_{0} = 36 cm$

$∴ |m_{D}| = \frac{v_{0}}{u_{0}} (1 + \frac{D}{f_{e}}) = \frac{36}{4.5} (1 + \frac{24}{8}) = 32$

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