Question

A microscope has a 1.7-cmcm-focal-length eyepiece and a 0.75-cmcm objective lens.

Part a: Assuming a relaxed normal eye, calculate the position of the object if the distance between the lenses is 14.8 cmcm .

Part b: Calculate the total magnification.

Answer 1

The focal length of eyepiece is fe = 1.7cm anf of objective lens fo = 0.75cm

the distance between the lenses are d = 14.8 cm , as fro relaxed eye we take the final image distance be 25 cm (near point of vision)

a) let the object distance be do , the image distance be di for the objective lens and

the object distance (image of the objective taken as object) from eyepiece be do' and the final image distance be di'

now using lens formula for eyepiece

putting the values
fe = + 1.7 cm , di' = 25 cm

we get do' = - 1.82 cm   ie 1.82 cm away from the eyepiece in between the lens ,

so di = d- do' = 14.8 - 1.82 = 12.17 cm

now using lens formula for eyepiece

putting the values
fe = + 0.75 cm , di = 12.17 cm

we get do = - 0.8 cm   ie 0.8 cm to the left of objective lens

so the object should be placed at 0.8 cm from the objective lens

b) the magnification of the lens is given by

putiing di' = 25 cm and do = 0.8 cm

we get m = 31.25