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Microscope to view sample in focus with a completely relaxed eye

by Niki4444
Tags: infinity, lenght, lens, microscope, physics
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Niki4444
#1
Jul20-08, 01:46 PM
P: 9
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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G01
#2
Jul20-08, 02:34 PM
HW Helper
G01's Avatar
P: 2,685
I don't see why you need to use magnification to solve this problem. I think your making this harder than it has to be.

The length of the tube, L, must be long enough to contain the distance of the first image from the first lens and the distance from the first image to the second lens. You know the distance of the first image from the first lens is 4.3cm and that the distance from the second lens has to be 2.5 cm. You should be able to find L just from this information.

If you haven't already, try drawing a picture to help you to visualize the setup of this microscope.
Niki4444
#3
Jul20-08, 02:36 PM
P: 9
Thank you for letting me know I'm making physics harder than it needs to be! I certainly don't want to do that. So L is the distance from the object to the final image??
So, you said.....
The length of the tube, L, must be long enough to contain the distance of the first image from the first lens and the distance from the first image to the second lens. You know the distance of the first image from the first lens is 4.3cm and that the distance from the second lens has to be 2.5 cm. You should be able to find L just from this information.

so L must be 4.3 cm + 2.5 cm??? I guess I don't understand what L is.

G01
#4
Jul20-08, 02:45 PM
HW Helper
G01's Avatar
P: 2,685
Microscope to view sample in focus with a completely relaxed eye

Quote Quote by Niki4444 View Post
Thank you for letting me know I'm making physics harder than it needs to be! I certainly don't want to do that. So L is the distance from the object to the final image??
So, you said.....
The length of the tube, L, must be long enough to contain the distance of the first image from the first lens and the distance from the first image to the second lens. You know the distance of the first image from the first lens is 4.3cm and that the distance from the second lens has to be 2.5 cm. You should be able to find L just from this information.

so L must be 4.3 cm + 2.5 cm??? I guess I don't understand what L is.

Yes. L is the length of the tube between the first and second lens. So if the image is inside the tube, 4.3cm away from one lens and 2.5 cm away from the other, then we can reason that the tube must be 2.5+4.3 = 6.8cm long.


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