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In a compound microscope the focal length of objective lens is 1.2 cm and focal length of eye piece is 3.0 cm .When object is kept at 1.25 cm in front of objective , final image is formed at infinity . Magnifying power of the compound microscope should be :

$(a)\;200\qquad(b)\;100\qquad(c)\;400\qquad(d)\;150$

1 Answer

Given $$f_e = 3 cm $$ and 
           $$f_o = 1.2 cm $$
           $$u_o = 1.25 cm $$
           $$v_e = \infty $$
From , $$\displaystyle \frac{1}{u_o} + \frac{1}{v_o} = \frac{1}{f_o} $$,
we get $$\displaystyle \frac{1}{v_o} = \frac{1}{1.2} - \frac{1}{1.25}$$
$$v_o = 30$$ cm 
Magnification of microscope $$\displaystyle m=\frac{v_o}{u_o} (1+\frac{D}{f_e})$$
$$\displaystyle  m = \frac{30}{1.25} (1+\frac{25}{3}) \simeq 200 $$            .

answered Apr 3, 2015 by ashu.gaikwad.56
 

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