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प्रश्न
The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focussed on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye piece.
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उत्तर
For the least distance of clear vision, the total magnification is given by:
`m=-L/`
where, L is the separation between the eyepiece and the objective
fo is the focal length of the objective
fe is the focal length of the eyepiece
D is the least distance for clear vision
Also, the given magnification for the eyepiece:
`m_e=5=(1+D/f_e)`
`=>5=1+20/f_e`
⇒fe= 5 cm
Substituting the value of m and me in equation (1), we get:
m=mo.me
`=>m_o=m/m_e=20/5=4`
Now, we have:
`m_0=L/(|f_0|)`
`=>f_0=14/4=3.5 "cm"`
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