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Question
A compound microscope consists of an objective of focal length 1 cm and an eyepiece of focal length 5 cm. An object is placed at a distance of 0.5 cm from the objective. What should be the separation between the lenses so that the microscope projects an inverted real image of the object on a screen 30 cm behind the eyepiece?
Solution
For the compound microscope, we have:
Focal length of the objective, f0 =1.0 cm Focal length of the eyepiece, fe = 5 cm Distance of the object from the objective, u0 =0.5 cm
Distance of the image from the eyepiece, ve =30 cm
The lens formula for the objective lens is given by
`1/v_0 -1/u_0 = 1/f_0`
`=> 1/v_0 +1/0.5 =1/1`
`=>1/v_0 =1 -10/5 =-1`
⇒ v0= -1 cm
The objective will form a virtual image at the side same as that of the object at a distance of 1 cm from the objective lens. The image formed by the objective will act as a virtual object for the eyepiece.
The lens formula for the eyepiece is given by
`1/v_e -1/u_e =1/f_e`
`=> 1/30 -1/u_e =1/5`
`=> -1/u_e =1/5 -1/30 =(6-1)/30 =1/6`
⇒ ue = -6 cm
∴ Separation between the objective and the eyepiece = 6 − 1 = 5 cm
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