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Question
A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope.
Solution
First we shall find the image distance for the objective`(v_0)`,
`1/f_0 = 1/v_0 -1/u_0 ; f_0 = 4cm,u_0 =-6cm`
`=> v_0 =12 cm`
Magnification of the microscope is,
`m = m_0 m_e = (v_0)/(u_0)(1+D/f_e) = (12/-6)(1 +25/10)`
= − 7, negative sign indicates that the image is inverted.
The length of the microscope is vo+ u, u=|ue| is the object distance for the eyepiece. And ue can be found using,
`1/f =1/D - 1/u_e`; as D is the image distance for the eyepiece.
`=> 1/10 =1/-25 - 1/u_e => u_e = -7.14 cm`
Hence, u = |ue| = 7.14 cm.
Length of the microscope vo+ u= 19.14 cm
Length of the microscope is given as
`L =(mf_0f_e)/D = (7 xx 4 xx 10)/25 = 11.2 cm`
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