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A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eyeplece of focal length 2.5 cm can bring an object placed at 2.0 mm from the objective in sharp focus. What is the separation between the two lenses ? Calculate the magnifying power of the microscope. |
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Answer» Solution :Angular MAGNIFICATION of the eye piece for IMAGE at 25 cm `=(1+(D)/(f))=1+(25)/(2.5)=11` for objective, `u=-0.9cm, f=0.8cm` `(1)/(UPSILON)-(1)/(u)=(1)/(f)` `(1)/(upsilon)=(1)/(0.8)-(1)/(0.9)=(9-8)/(7.2)=(1)/(7.2)` `rArr upsilon =7.2cm.` for eye piece, `upsilon = -25CM, f_(0)=25cm` Again `(1)/(upsilon)-(1)/(u)=(1)/(f)` `(-1)/(u)=(1)/(f)-(1)/(v)=(1)/(2.5)-(1)/(-25)=(11)/(25)` `u=-(25)/(11)=-2.27cm` `therefore` Distance between objective and eye piece `=v+|u|=7.2+2.27=9.47cm` MAGNIFYING power of microscope `=(7.2)/(0.9)xx(-25)/((-25)/(1))=88` |
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