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(a) Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision. (b) The total magnification produced by a compound microscope is 20. The magnification produced by the eyepiece 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 eyepiece. |
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Answer» SOLUTION :(a) N/A (b) (b) Here total magnification m = 20, magnification of EYEPIECE me = 5, distance between objective and eyepiece L = 14 CM, least distance of distinct vision D = 20 cm. magnificationof eyepiece `m_(e) =v_(e)/u_(e) =D/u_(e) rArr u_(e) =D/m_( e) = (20 cm)/5 = 4 cm` `therefore` Focal length of eyepiece `1/f_(e) =1/v_(e)-1/u_(e) =1/(-20) -1/(-4) =1/4 -1/20 =1/5 rArr f_(e) = 5 cm` `therefore` Focal length of objective `1/f_(0) =1/v_(0) -1/u_(0) =1/10-1/(-2.5) = 1/2 rArr f_(0)=+ 2cm` |
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