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A compound microscope consists of an obj...

A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (a) the least distance of distinct vision (25cm), and (b) at infinity? What is the magnifying power of the microscope in each case ?

Text Solution

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a. For the eyepiece `v_e=-25cm nd f_e=+5cm`
Using `1/v_e-1/u_e=1/f_e`
`1/u_e=1/v_e-1/f_e`
`-1/(25cm_-1/(5cm)`
or `u_e=25/6cm=-4.17=-4.2cm`
As the objective is 12.2 cm away from the eyepiece the image formed by the objective is 12.2 cm-4.2 cm =8.0 cm away from it. for the objective
`v=+8.0 cm, f_0=+1.0cm`
Using `1/v-1/u=1/f_0`
`1/u=1/v-1/f_0`
`=1/(8.0cm)-1/(1.0cm)`
or `u=-8.0/7.0cm=-1.1cm.`
b. The angular magnification is
`m=v/u(1+D/f_e)`
`=(+8.0cm)/(-1.1cm)(1+(25cm)/(5cm))=-44`.
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