What is the magnifying power of the microscope if the eye piece and the image is formed at 25 cm from the eye piece?In a compound microscope the focal length of the objective and the eye piece are 0.5cm and 2.5cm respectively. The real image formed by the objective is 16 cm from it ?
Answers
The magnifying power of a compound microscope is the product of the linear magnification of the objective and the magnifying power of the eyepiece.
⇒M.P.=Mo×Me=(vouo)(Due), where,
u,v and D are the distances of the object, image and the distance of distinct vision, respectively, and,
The subscripts o and e stand for the object and eyepiece respectively.
For the objective lens, 1vo−1uo=1fo⇒vo=uofouo+fo.
⇒vouo=fouo+fo.
⇒M.P.=(vouo)(Due)=(fouo+fo)(Due).
As per the new Cartesian convention, uo,ue and D are negative while vo and fo are positive.
Further, |uo|>|fo|.
Hence, taking absolute values of the distances, we get,
M.P.=−(fouo−fo)(Due).
It is given that fo=1 cm,fe=5 cm and uo=1.1 cm.
The distance of distinct vision, D, may be taken as 25 cm.
⇒M.P.=−(fouo−fo)(Due)=−(11.1−1)(255)=−50.
⇒ The magnifying power of the combination of lenses is −50.
The negative sign indicates that the image is inverted.