Question

The distance between the objective and the eyepiece of a compound microscope is 64 mm. The focal length of the objective is 8 mm, and the object distance is 12 mm. The overall magnification is -12. What is the focal length (in mm) of the eyepiece?

Answer 1

See the diagram :

Given :

fo = focal length of objective lens = 8 mm

u = object distance = 12 mm

L = distance between the lenses = 64 mm

Overall Magnification = M = -12

fe = focal length of eyepiece = to be calculated.

I will use cartesian sign convention, in which the measurement taken in the direction of incident light is positive and opposite to incident ray is negative.

Applying Lens's equation for objective :

[u = object distance, v = image distance, f = focal length]

[vo = image distance from objective]

. [the image formed is real and inverted]

The image formed by objective will behave as object for eyepiece.

ue = object distance for eyepiece = L - vo = 64 - 24 = 40 mm.

Also, Total Magnification = M = Mo x Me [Mo = magnification by objective, Me = magnification by eyepiece]

mm [-ve sign mean the image formed by eyepiece is virtual]

Applying Lens's equation for eyepiece :

[u = object distance, v = image distance, f = focal length]

[ve = image distance from objective]

focal length of eyepiece is 48 mm.