Two lenses are mounted in a tube such that the distance between them can be vari

Question

Two lenses are mounted in a tube such that the distance between them can be varied. This device can be used as either a telescope or a compound microscope. One lens (the objective) has a focal length of 51 cm.

Find the focal length of the second lens (the eyepiece) such that the magnification |m|=si sN/(fobj feyepiece) of the microscope is the same as the magnification |m|=fobj/feyepiece of the telescope. Assume the distance between the two lenses of the microscope (si+feyepiece) is exactly twice as large as the distance between the two lenses of the telescope. For the near-point distance use sN=25 cm.

Explanation / Answer

We know that length of telescope = fo+fe

So given is

Si+ fe= 2(fo+fe) => Si= 2fo+ fe

And SN = 25 cm

So for magnification to be equal

(2fo+ fe)*25/(fo*fe) = f0/fe

Given fo= 51& assume fe=x

( 102+x)*25 /51x = 51/x

=> 102 +x = 51*51x/25x => x = 2.04m

So m = 51/2.04 = 25

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