Question

A microscope has an objective lens with f = 16.22 mm and an eyepiece with f = 9.5 mm. With the length of the microscope's barrel set at 29.0 cm, the diameter of an erythrocyte's image subtends an angle of 1.43 mrad with the eye. If the final image distance is 29.0 cm from the eyepiece, what is the actual diameter of the erythrocyte?

Answer 1

given
f = 16.22mm
focal length of eye piece = 9.5mm
length of barrel set=29cm
angle = 1.43 m rad
image distance = 29cm
m =[ L / fo ] * [ 1 +D / fe]

Where D is the distance of distinct vision = 270 mm
Using the given values

m = 270 mm / 16.22 mm { 1 + 270 mm / 9.5 mm}

m = 489.74= size of the image / size of the object.

489.74=( 29 cm * 1.43 x 10^-3 ) / ( size of the object.)

size of the object = 29 cm x 1.43 x 10^-3 / 489.74=0.0846*10^-3 cm

=84.6μ m