InterviewSolution
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InterviewSolution
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A compound microscope uses an orjective lens of focal length 4 cm and eyepiece lens of focal Length 10cm. AN object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the mircroscope. |
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Answer» Solution :`(1)/(v_(0))- (1)/(u_(0)) = ( 1)/(f_(0))` `f_(0) = 4 cm, u_(0) = - 6 cm` `(1)/(v_(0)) = (1)/(4) - (1)/(6) = (1)/(12) rArr v_(0) = 12 cm` magnification by objective, `m_(0) = (v_(0))/(|u_(0)|)` `= (12)/(6) = 2 ` Magnification by EYEPIECE `m_(E) = ( 1+ ( D)/( f_(e))) ` or `( D)/( f_(e))` `= ( 1+ ( 25)/( 10)) ` or `(25)/(10)` magnification power of the microscope `m= m _(0) xx m_(e )` `= 2 xx 3.5 ` or `2 xx 2.5 ` =7 or 5 Length of the microscope , `L = | v_(0) | +| u_(0)|` `L= |v_(0)|+f_(e)` `u_(e ) = ?, v_(e) = D = - 25 cm, f_(e ) = 10CM` `(1)/(u_(e )) = ( 1)/( v_(e)) - (1)/(f_(e))= - (1)/(25) - ( 1)/(10) = - (7)/( 50)` `u_(e ) = - ( 50)/( 7) cm` `:. L = 12 + ( 50)/( 7)` = 19.1 cm |
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