(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A microscope has a converging lens (eyepiece) with focal length of 2.50 cm mounted on one end of a tube of adjustable length. At the other end is another converging lens (objective) with a focal length of 1.00 cm. When you place the sample 1.30 cm from the objective, what length, l, will you need to adjust the tube of the microscope to view the sample in focus with a completely relaxed eye. It then explains that to view sample with a completely relaxed eye, the eyepiece must form its image at infinity.

2. Relevant equations

f(eyepiece)=2.50 cm

f(objective)=1.00 cm

L=adjustable

1/f=1/s+1/s'

Mo=-s'/s

Me=25/f(eyepiece)

M=(Mo)(Me)=-(L/fo)(25 cm/f(eyepiece))

3. The attempt at a solution

so, for the objective lens,

1/f=1/s+1/s'

1/1=1/1.3+1/s'

solve for s' and get 4.3 cm

so, for the eyepiece lens,

1/f=1/s+1/s'

1/2.5=1/2.5+1/infinity

so, I thought s would be close to 2.5

so, Mo=-(4.3)/(1.3)=-3.3

so, Me=25/2.5=10

so, M=(Mo)(Me)=(-3.3)(10)=-33.1

M=-33.1=(-L)/(fo)*(25/fe)

-33.1=-L/1*25/2.5

solving for L, I get 3.31 cm which is incorrect

Any ideas to push me toward a better solution? Thanks again for your help.

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# Microscope to view sample in focus with a completely relaxed eye

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