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Q.

A compound microscope has a magnification of 30. Assuming that the final image is formed at the least distance of distinct vision (25 cm), find the magnification produced by the objective. Given, the focal length of the eyepiece is 5.0 cm.

OR

A refracting (astronomical) telescope, when in normal adjustment, has a magnifying power of 6 and the objective and the eyepiece are 14 cm apart. Find the focal lengths of the lenses of the telescope.

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Detailed Solution

Magnification, M=30

Focal length, f=5 cm

Final distance, D=25 cm

 In a compound microscope, M=mo×me

Thus,
30=mo1+Df 30=m01+255  Thus, m0=5
OR
Given,
magnification=6,
length=14. Length of the telescope =fo+fe ......(1)  Magnification =fofe=6  .....(2) From 1 and 2 fe=2cm f0=12cm

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A compound microscope has a magnification of 30. Assuming that the final image is formed at the least distance of distinct vision (25 cm), find the magnification produced by the objective. Given, the focal length of the eyepiece is 5.0 cm.ORA refracting (astronomical) telescope, when in normal adjustment, has a magnifying power of 6 and the objective and the eyepiece are 14 cm apart. Find the focal lengths of the lenses of the telescope.