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An angular magnification (magnifying power) of 30 X is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5cm. How will you set up the compound microscope ? |
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Answer» Solution :In normal ajustment, image is formed at least distance of distinet vision, d = 25 cm Angular MAGNIFICATION of EYE piece `=(1+(d)/(f_(e)))=(1+(25)/(5))=6` As total Magnification is 30, Magnification of objective lens. `m=(30)/(6)=5` `m=(upsilon_(0))/(-u_(0))=5, or v_(0)=-5u_(0)` As `(1)/(v_(0))-(1)/(u_(0))=(1)/(f_(0))` `(1)/(-5u_(0))-(1)/(u_(0))=(1)/(1.25)` `(6)/(-5u_(0))=(1)/(1.25)` `u_(0)==(-6xx1.25)/(5)=-1.5cm.` i.e. object should be held at 1.5 cm in - front of objective lens As `v_(0)=-5u_(0)` `v_(0)=-5(-1.5)=7.5cm` From `(1)/(upsilon_(e))-(1)/(uu_(2))=(1)/(f_(e))` `(1)/(u_(e))=(1)/(v_(e))-(1)/(f_(e))` `=-(1)/(25)-(1)/(5)=-6//25` `u_(e)=(-25)/(6)=-4.17cm` `therefore` Seperation between the objective lens and eye piece `|u_(e)|+|v_(0)|` `=4.17+7.5` `=11.67cm` |
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