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CBSE Class 12 Physics 2014 Delhi Set 1 Paper

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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  cm14\;\mathrm{cm}. If least distance of distinct vision is 20  cm20\;\mathrm{cm}, calculate the focal length of the objective and the eye piece.
Solution:  
(i)
(ii) From m  =m0×mem\;=m_0 \times m_e
m0  =mme=205=4m_0\;=\frac{m}{m_e}=\frac{20}{5}=4
Now, m0  =v0u0=Lf0=4m_0\;=\frac{v_0}{u_0}=\frac{L}{f_0}=4
f0  =L4=144=3.5  cmf_0\;=\frac{L}{4}=\frac{14}{4}=3.5\;\mathrm{cm}
Also, me  =1+dfe=5m_e\;=1+\frac{d}{f_e}=5
dfe  =5−1=4\frac{d}{f_e}\;=5-1=4
fe  =d4=204=5  cmf_e\;=\frac{d}{4}=\frac{20}{4}=5\;\mathrm{cm}
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