The focal lengths of the objective and the eyepiece of a compound microscope are 4cm and 6 cm respectively. If object is placed at distance 6 cm infront of objective lens, then maximum magnification of microscope is 
1.) 31/3
2.) 62/3
3.)15/3
4.)10/3
Please explain.

Asked by Kb Aulakh | 25th Apr, 2015, 05:46: PM

Expert Answer:

Given:
uo=-6 cm, fo=4 cm, fe= 6 cm, D = 25 cm
begin mathsize 14px style The space image space of space the space object space is space formed space by space the space object space lens space at space distance space straight v subscript straight o space. space The space distance space straight v subscript straight o space is space given space by comma
1 over straight f subscript straight o equals 1 over straight v subscript straight o minus 1 over straight u subscript straight o
1 over straight v subscript straight o equals 1 over straight f subscript straight o plus 1 over straight u subscript straight o
space space space space space space space equals 1 fourth minus 1 over 6
space space space space space space space equals fraction numerator 6 minus 4 over denominator 24 end fraction equals 1 over 12
straight v subscript straight o equals 12 space cm
Magnification comma space straight M space equals space straight v subscript straight o over straight u subscript straight o left parenthesis 1 plus straight D over straight f subscript straight e right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 12 over denominator negative 6 end fraction left parenthesis 1 plus 25 over 6 right parenthesis
space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 12 over denominator negative 6 end fraction left parenthesis fraction numerator 6 plus 25 over denominator 6 end fraction right parenthesis space
space space space space space space space space space space space space space space space space space space space space space space space space space space space equals space fraction numerator 12 over denominator negative 6 end fraction left parenthesis 31 over 6 right parenthesis space equals space fraction numerator 12 space cross times 31 over denominator negative 36 end fraction space equals space minus 31 over 3 equals negative 10.33 end style

Answered by Faiza Lambe | 25th Apr, 2015, 10:16: PM