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Draw a ray diagram to show the working of a compound microscope . Deduce an expression for the total magnification when the final is formed at the near point . In a compound microscope, an object is place at a distance of 1.5cm from the objective of focal length 1.25cm. If the eye piece has a focal length of 5cm and the final image is formed atthe near point, estimate the magnifying power of the microscope. |
Answer» Solution :Labelled Ray DIAGRAM `:`  Expression for total magnification `:` Magnification due to the objective , `m_(0) = ( h.)/( h ) = ( L)/( f_(0))` Magnification `m_(E )`, due to eyepiece , ( when the final image is formed at the NEAR point ) `m_(e ) = ( 1+ ( D)/( f_(e )))` Total magnification, `m= m_(0) m_(e ) = ( L)/( f_(0)) ( 1+ (D )/( f_(e)))` Estimation of magnifying POWER `:` Given `:` `u_(0) = - 1.5 cm, f_(0) = 1.25 cm`, we have `(1)/( f_(0)) = ( 1)/( v_(0)) - ( 1)/( u _(0))` `(1)/(1.25) = ( 1)/( v_(0)) - ( 1)/( - 1.5) rArr v_(0) = 7.5 cm` `m= (v_(0))/( u_(0)) ( 1 + ( D)/( f_(e)))` `= ( 7.5 )/( - 1.5 ) ( 1+ ( 25)/( 5)) rArr m = - 30` |
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