CBSE Class 12 Physics 2014 Delhi Set 1 Paper

Question : 22 of 30

Marks: +1, -0

(i) Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision.
(ii) 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 eye-piece 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.

Solution:

(i)

(ii) From

m\;=m_0 × m_e

m_0\;=\;{m}/{m_e}=\;{20}/{5}=4

Now,

m_0\;=\;{v_0}/{u_0}=\;{L}/{f_0}=4

f_0\;=\;{L}/{4}=\;{14}/{4}=3.5 {\cm}

Also,

m_e\;=1+\;{d}/{f_e}=5

\;{d}/{f_e}\;=5-1=4

f_e\;=\;{d}/{4}=\;{20}/{4}=5 {\cm}

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