The magnifying power of a microscope with an objective of 5 mm focal length is 400. The length of its tube is 20 cm. Then the focal length of the eye-piece is (consider that final image is formed at infinity)

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
$m_{\infty} = \frac{(L - f_{o} - f_{e}) D}{f_{o} f_{e}} ≃ \frac{L D}{f_{o} f_{e}}$

$\Rightarrow 400 = \frac{20 \times 25}{0.5 \times f_{e}} \Rightarrow f_{e} = 2.5 c m .$

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The magnifying power of a microscope with an objective of 5 mm focal length is 400. The length of its tube is 20 cm. Then the focal length of the eye-piece is [MP PMT 1991]

Q. The magnifying power of a microscope with an objective of 5 mm focal length is 400. The length of its tube is 20 cm. Then the focal length of the eye-piece is (consider that final image is formed at infinity)

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Q. In a compound microscope the focal length of objective lens is

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. When object is kept at

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in from of objective, final image is formed at infinity. Magnifying power of the compound microscope should be :

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The magnifying power of a microscope with an objective of 5 mm focal length is 400. The length of its tube is 20 cm. Then the focal length of the eye-piece is (consider that final image is formed at infinity)

The correct option is C 2.5 cmm∞=(L−fo−fe)Dfofe≃LDfofe ⇒400=20×250.5×fe⇒fe=2.5cm.

The correct option is C 2.5 cm
$m_{\infty} = \frac{(L - f_{o} - f_{e}) D}{f_{o} f_{e}} ≃ \frac{L D}{f_{o} f_{e}}$

$\Rightarrow 400 = \frac{20 \times 25}{0.5 \times f_{e}} \Rightarrow f_{e} = 2.5 c m .$