The magnifying power of a microscope with an objective lens of 5 mm focal length is 400. The length of its tube is 20 cm. Then the focal length of the eyepiece is

200 cm

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160 cm

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2.5 cm

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0.1 cm

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Solution

The correct option is C 2.5 cm
$M_{\infty} = \frac{L D}{f_{0} f_{e}} 400 = \frac{20 \times 25}{0.5 \times f_{e}} \Rightarrow f_{e} = \frac{/ 20 \times 25}{/ 400 20 \times 0.5} f_{e} = 2.5 c m$

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The magnifying power of a microscope with an objective lens of 5 mm focal length is 400. The length of its tube is 20 cm. Then the focal length of the eyepiece is

The correct option is C 2.5 cmM∞=LDf0fe400=20×250.5×fe⇒fe=/20×25/40020×0.5fe=2.5 cm

The correct option is C 2.5 cm
$M_{\infty} = \frac{L D}{f_{0} f_{e}} 400 = \frac{20 \times 25}{0.5 \times f_{e}} \Rightarrow f_{e} = \frac{/ 20 \times 25}{/ 400 20 \times 0.5} f_{e} = 2.5 c m$