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The total magnification produced by a co...

The total magnification produced by a compound microscope is `20`. The magnification produced by the eye piece is `5`. The microscope is focussed on a certain object. The distance between the objective and eye piece is observed to be `14 cm`. If least distance of distinct vision is `20 cm`, calculate the focal length of objective and eye piece.

Text Solution

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From `m = m_(0) xx m_(e)`,
`m_(0) = (m)/(m_(e)) = (20)/(5) = 4`
Now, `m_(0) = (v_(0))/(u_(0)) = (L)/(f_(0)) = 4`,
`f_(0) = (L)/(4) = (14)/(4) = 3.5 cm`
Also, `m_(e) = 1 + (d)/(f_(e)) = 5`
`(d)/(f_(e)) = 5-1 = 4, f_(e) = (d)/(4) = (20)/(4) = 5 cm`
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