A compound microscope in which the objective and eyepiece are17 cm apart uses an

Question

A compound microscope in which the objective and eyepiece are17 cm apart uses an objective lens with f(objective) = 0.7 cm andan eyepiece with f(eyepiece) = 5.3 cm. How wouldthe magnification change if the focal length of the objective lenswere decreased by a factor of two?   
I know the magnification of the microscope would increase butI am having difficulties figuring out by what factor. Please help. A compound microscope in which the objective and eyepiece are17 cm apart uses an objective lens with f(objective) = 0.7 cm andan eyepiece with f(eyepiece) = 5.3 cm. How wouldthe magnification change if the focal length of the objective lenswere decreased by a factor of two?   
I know the magnification of the microscope would increase butI am having difficulties figuring out by what factor. Please help.

Explanation / Answer

Magnification of compound microscoope = (L*D)/(fo*fe) L = Tube length of the compound microscope which is thedistance between the 2nd focal point of the objective and the 1stfocal point of the Eye piece. L = 17-0.7-5.3 = 11 cm D is the least distance of distinct vision which is usuallytaken as 25 cm fo = 0.7 cm fe = 5.3 cm Magnification = (11*25)/(0.7*5.3) = 74.124 Now fo is decresed by a factor of 2, so fo = 0.7/2 = 0.35cn Now = 17-0.35-5.3 = 11.35 cm so new magnification = (11.35*25)/(0.35*5.3) = 152.965 So the reduction of focal length by factor of 2 willincrese the magnification by 152.965/74.124 = 2.064 Almost incresed by a factor of 2.

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