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Question
The separation between the objective and the eyepiece of a compound microscope can be adjusted between 9.8 cm to 11.8 cm. If the focal lengths of the objective and the eyepiece are 1.0 cm and 6 cm respectively, find the range of the magnifying power if the image is always needed at 24 cm from the eye
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Solution
For the compound microscope, we have:
Focal length of the objective, fo = 1.0 cm
Focal length of the eyepiece, fe = 6 cm
Image distance from the eyepiece, ve = -24 cm
Separation between the objective and the eyepiece = 9.8 cm to 11.8 cm
The lens formula is given by
`1/v_e - 1/u_e = 1/f_e`
`1/-24 - 1/u_e =1/6`
`=> 1/u_e = -1/6-1/24`
`=>1/u_e =-(4+1)/24 = -5/24`
`therefore u_e =-24/5 cm =-4.8 cm`
(a) Separation between the objective and the eyepiece = 9.8 cm
So, for the objective lens, the image distance will be
vo = 9.8 − 4.8 = 5 cm
The lens formula for the objective lens is given by
`1/v_0 -1/u_0 =1/f_0`
`1/5 -1/u_0 =1/1`
`=> -1/u_0=1-1/5 =(5-1)/5`
`u_0 =-5/4 cm =-1.25 cm`
Magnifying power of the compound microscope:
m=`v_0/u_0(1+D/f_e)`
=`5/1.25 (1+24/6 )=20`
(b) Separation between the objective and the eyepiece = 11.8 cm
We have:
m = 30
So, the required range of the magnifying power is 20–30.
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