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molip790

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A Homemade compound microscope has, as objective and eyepiece, thin lenses of focal lengths 1cm and 3cm, respectively. An object is situated at a distance of 1.20cm from the objective. If the virtual image produced by the eyepiece is 25cm from the eye, compute (a.) the magnifying power of the microscope and (b) the separation of the lenses

2. Homework Equations

M = (-25)(L)/(fo)(fe) where L is the length, and fo = focal length of objective, and fe = focal length of the eyepiece

1/s + 1/s' = 1/fo

M = 25/feff where feff is the effective focal length of the two lenses separated by a distance d

So... 1/feff = 1/fo + 1/fe - d/fofe

3. The Attempt at a Solution

(a.) 1/s + 1/s' = 1/fo

1/1.20 + 1/s' = 1/3

1/s' = -1/2

s' = -2cm

then...

M = -25L/fofe

= -25(s'-fo)/fofe

= -25(-2-3)/(3*1)

= 125/3

M = 41.6667 ? which isn't right (its supposed to be 46.7x)

which will make part b. wrong, but even with the correct answer using the equation I have:

M = 25/feff and feff = 1/fo + 1/fe - d/fofe I still get d = 2.39 which is also incorrect, its supposed to be 8.68