A microscope has an objective lens with f = 16.22 mm and an eyepiece with f = 9.

Question

A microscope has an objective lens with f = 16.22 mm and an eyepiece with f = 9.2 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 14.3 mrad with the eye. If the final image distance is 27.0 cm from the eyepiece, what is the actual diameter (in µm) of the erythrocyte? Hint: Start with the size of the final image, then use the thin-lens formula for each lens to find their combined magnification. Use this magnification to calculate the object size in the final step

Explanation / Answer

m =[ L / fo ] * [ 1 +D / fe]

Where D is the distance of distinct vision = 290 mm
Using the given values

m = 290 mm / 16.22 mm { 1 + 290 mm / 9.2 mm}

m = 581.46= size of the image / size of the object.

581.46=( 27 cm * 14.3 x 10^-3 ) / ( size of the object.)

size of the object = 27 cm x 14.3 x 10^-3 / 581.46= 6.640e-5 cm

=0. 764297 m

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