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The focallengths of the objective and the eyepiece of a compound microscope are 2.0cm and 2.0cm, respectively. The distance between the objective and the eyepiece is 15.0cm. Th final image formed by the eyepiece is at infinity. The two lenses are thin. The distance, in cm, of the object and the image produced by the objective, mesured from the objective lens, are respectively. |
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Answer» 2.4 and 12.0 Using lens FORMULS for eyepiece, `-(1)/(u)+(1)/(v_(1))=(1)/(f_(e))` `RARR (-1)/(u)+(1)/(prop)=(1)/(3)rArr u_(1)=-3cm[:'I=0]` But the distance between objective and eyepiece is 15cm (given). Therefore, distance of image formed by the objective, `v=15-3=12cm` . LET u be the object distance from the objective, then for objective lens `-(1)/(u)+(1)/(v)=(1)/(f_(0))or (-1)/(u)+(1)/(12)=(1)/(2)` `rArr (-1)/(u)=(1)/(2)-(1)/(12)=(5)/(12)u=-(12)/(4)=2.4cm` (a) is the CORRECT OPTION. |
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