Find the resolving power of a microscope in terms of its diameter

Tags:
1. Mar 19, 2013

mew1033

Find the resolving power of a microscope lens in terms of its diameter

1. The problem statement, all variables and given/known data
In this problem, we will find the ultimate resolving power of a microscope. First of all, in order to obtain a large magnification, we want an objective lens with a very short focal length. Second, in order to obtain maximum resolution, we also want that lens to have as large a diameter as possible. These two requirements are conflicting, since a lens with a short focal length must have a small diameter. It is not practical for a lens to have a diameter much larger than the radius of curvature of its surfaces. Otherwise, the lens starts looking like a sphere. So, let us assume that the objective lens has a diameter D equal to the radius of curvature of the two surfaces, like the lens in the figure.
(a) If the lens is made of glass with index of refraction 1.54, find the focal length f in terms of the diameter D of the lens.
(b) The distance between the sample to be observed and the objective lens is approximately equal to the focal length f . Find the distance between two points on the sample which can be barely resolved by the lens. Use the result from part (a) to eliminate f from the expression. You should find that D is
also eliminated from the expression and that the answer is given entirely in terms of the wavelength λ of the light. You may use the small angle approximation, sinθ ≈ tanθ ≈ θ.

The answer for part a is given as being in the range 0.8-1.3 $D$
The answer for part b is given as being in the range 1.0-1.6$\lambda$
2. Relevant equations
I think that we will use the lens makers equation: $1/f=(n-1)(1/R_1-1/R_2)$ for part a. Then for part b, I think it's Rayleigh's criterion: $sin\theta=\lambda/a$

3. The attempt at a solution

I'm completely stuck at part a... I'm not really sure what I should use for $R_1$ and $R_2$.
I think if I got part a, part b would make more sense.

Thanks

Attached Files:

• Figure.png
File size:
574 bytes
Views:
128
Last edited: Mar 19, 2013
2. Mar 19, 2013

rude man

"So, let us assume that the objective lens has a diameter D equal to the radius of curvature of the two surfaces."

3. Mar 19, 2013

mew1033

Do you mean set D equal to R1 and R2? If I do that, then (1/R1-1/R2) comes out to be 0. Then the focal length is basically infinity.

4. Mar 19, 2013

MrAnchovy

What is the sign convetion for $R_1$ and $R_2$ in the equation you are using?

5. Mar 19, 2013

mew1033

I'm not sure what you mean by that, sorry....

I talked to another student and they said that when you use the lens-makers equation, if it's a converging lens then you do $1/f=(n-1)(1/R_1+1/R_2)$

Looks like that was my problem.

6. Mar 20, 2013

rude man

That's the formula I would use. If it's convex to the outside it's +. So both faces are convex to the outside and so it's 1/R1 + 1/R2. It obviously makes sense too. A double-convex lens is obviously not equivalent to a flat piece of glass!

Some of these optial sign conventions seem bizarre but to be candid if you don't follow them sooner or later you get into trouble. Especially with object/image/virtual/real blah blah problems.