Mathematics of Microscope Resolving Power

Mathematics of Microscope Resolving Power

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Missing Figures

Imagine this, you are walking through the forest when all of a sudden you come across the most fascinating insect (perhaps insects may not seem too fascinating at first but once you learn a little about them they are the most fascinating creatures). Well, back to the story, so you find this insect and you realize that it seems very different from those you've previously encountered. Well, being the curious scientist that you are, you take out your trusty magnifying glass and take a look. You move the lens back and forth until you find the perfect image. You see the insect's wonderful colours and patterns which you would not be able to see with your naked eye. What just happened? You simply placed a piece of glass between you and the insect and all of a sudden you get this wonderful view of nature which would otherwise be missed. Well, if you are at all curious as to know how magnifying glasses and microscopes work, then read on and find out.

An Introduction to Microscopes

The two types of microscopes that will be focused on in this webpage are the simple microscope and the compound microscope. The simple microscope, also known as the magnifying glass, is composed of a single converging lens. The compound microscope is composed of at least two lenses and is generally referred to as a microscope.

There are two main purposes of a microscope:


1) to increase the magnification of an object
2) to have a high resolving power

Both of these will be examined; however, a greater emphasis will be placed on the resolving power.

Magnifying Power (brief overview)

Magnifying power: is also called angular magnification. Figure 1a shows an object y in front of a lens. Rays of light reflect off the object through the lens and a now larger image, y', of y can be seen. Once, the image is brought further from the lens, as in figure 1b, the image, y', is even larger. (So as to no discrepency: in figures 1a and 1b, the observer is on the right of the lens looking towards the image y')

The magnifying power, M, is given by the following:
M = 1 + d/f, where f is the focal distance and d is the distance between the object and the lens

Proof of M = 1 + d/f:


Figure 1c is the view of the object Y from point C without a magnifying glass.

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Figure 1d is the view of the object y from point E with a magnifying glass.

let:



* M = magnification
* y, Y = object
* D = distance between object and lens (within focal length - if beyond focal length you get a virtual image)
* y' = image
* d = distance between image and lens
* f = focal length



by definition, M = u/v
Since u and v are small, let u = tanu and v = tanv
Thus M = tanu / tanv (1)
In the figure 1c, triangle ABC is right; thus tanv = Y/D (2)
Right triangle HIE gives tanu = y/d
Since y = Y, tanu = Y/d (3)
Thus, substituting (2) and (3) into (1) you get M = (Y/d) / (Y/D) = D/d
Substituting into the thin lens formula: 1/f = -1/D + 1/d
Multiplying both sides by D: D/f = -1 + D/d, and adding one to both sides: D/d = 1 + D/f = M




Total magnification can be found by: M1 x M2 = M(total), where M1 and M2 are magnifications
Thus, in a compound microscope (a microscope with more than one lens) the magnification can be found by:
M(total)= M1 x M2 where M1 = -L/f(o) = magnification of the objective, M2 = 1 + D/f(e) = magnification of eyepiece (f(o)= focal point of objective, f(e) = focal point of eye piece, L = distance between the two lenses, D = distance between object and first lens)
Thus M = (-L/f(o)) + (1 + D/f(e))*
*The negative sign means the image seen is upside down.

Notes:

* The maximum magnification that can be seen with a light microscope is 2,000x

* The maximum magnification that can be seen with an electron microscope is 200,000x

Resolving Power (Page 1)

Numerical aperture, NA, is the "half-angle subtended at the object by the entrance pupil of the system (usually the first lens) and the index of the viewing medium" (Blaker, Geometric Optics the Matrix Theory).
Allowing n = refractive index of the object, and u = angle of the cone of light coming from the objective, the numerical aperture can be found by NA = n sinu.

What is the largest NA?
NA = nsinu; thus, we want the largest n and largest sinu. As will be explained later, the maximum sinu is 0.95. The index of refraction in air is 1.0 and the index of refraction of oil and glass cover (ie lens dipped in oil) is 1.515. Thus, using n = 1.515, the "largest" NA can be found:
NA = (1.515)(0.95) = 1.44

The limiting angle of resolution, u(min), with an objective diameter of b is given by the equation:
u(min) = 1.22? / b

"Proof" of u(min) = 1.22? / b:
let:



* a = width of the slit
* u = angle of deflection
* u(min) = angle of limiting resolution
* b = diameter of lens
* ? = wavelength


As seen in figure 7*, ray1 travels a/2 sinu further than ray 3 (ray 2 and ray 3 travel a/2 sinu further than ray 4 and ray 5 respectively)



There is destructive interference (the viewing screen is dark) when: a/2 sinu = ? / 2
Therefore, sin u = ? / a
However, ? is very small (around 5*10^(-9)) so u is approximately equal to sinu. Thus, it can be written: u(min) = ? / a
However, microscopes are not slits (they are circulare apertures) so u(min) must be greater if the image is to be resolved.
Many experiments have been performed to find out just how much larger this u(min) must be, their results showed that it must be 1.22 times larger. Thus, replacing the width of the slit with the diameter of the lens and increasing u(min) by a factor of 1.22, we get:
u(min) = 1.22 ? / b
*this figure is modelled after Figure 38.5 in Physics for Scientists and Engineers by Serway

The limiting angle of resolution can be decreased when:

* the diameter of the lens, b, is increased
* decreasing the wavelength, ? (ie by using filteres)

Note: Theoretically, the largest value of u should be 90 degrees (assuming the lens was large enough); however, in practice, the maximum view is only about 71.8 degrees (ie sinu = 0.95). Even though 71.8 degrees is the maximum angle that can be obtained, it is every difficult to achieve such a high u since equipment and environment must be ideal.

How can you minimize the limiting angle of resolution of a lens which has a diameter of 0.900cm?
By using the smallest visible light wave, which is 400nm, and immersing the object in oil (which has a refraction index, n, of about 1.5) the smallest limiting angle of resolution, u(min), can be found:
?(oil) = ?(air) / ?(oil) = 400nm / 1.5 = 268nm
u(min) = 1.22 ? / b = 1.22(0.0000267cm / 0.900cm) = 0.0000362 radians

Resolving power (Page 2)

Resolving power, or resolution, is the smallest distance between two separate points of an object, when viewed with an optical instrument, that can still be seen as distinguishable. A microscope's resolution limit, d, can be found by the following formula:
d = 0.61 ? / NA, where ? is the wavelength of light coming from the object, and NA is the numerical aperture. (This is called the Ernst Abbe formula)

"Proof" of d = 0.61 ? / NA:
let:

* ? = wavelength
* u = angle of the cone of light coming from object
* u' = angle of cone of light forming image
* n = refraction index of object
* m = magnification
* NA = numerical aperture
* d = distance between two points in the image

d = 0.61 ? / (m tanu') (1)* **
m = n sinu / sinu', thus m sinu' = n sinu = NA
However, tanu is approximately sinu when u is very small.
Therefore, m tanu' = NA (2)
Substituting (2) into (1): d = 0.61 ? / NA

*this formula is derived in "Theory of Optical instruments"
** all = not strict equalities but approximations
Note: A detailed proof of this formula can be found in "Theory of Optical Instruments"pp53-54, 69 & "Geometerical Optics" pp99-100, 125. However, to understand these proofs a high level of Mathematics is required

The resolving power increases when d, the minimum distance that can be seen between two points in the image, decreases. Thus, according to the formula d = 0.61 ? / NA, the resolving power can be increased in two ways:

* decreasing the wavelength, ? (ie by using filters)
* increasing the NA. As stated earlier, NA = n sinu. Thus, NA can be increased the following ways:
o increasing the refraction index, n (this can be done by adding oil to the object)
o increasing the angle of light coming from the object, u

The limiting angle of resolution can be decreased when:

* the diameter of the lens, b, is increased
* decreasing the wavelength, ? (ie by using filteres)

Note: Theoretically, the largest value of u should be 90 degrees (assuming the lens was large enough); however, in practice, the maximum view is only about 71.8 degrees (ie sinu = 0.95). Even though 71.8 degrees is the maximum angle that can be obtained, it is every difficult to achieve such a high u since equipment and environment must be ideal.

What is the limit of resolving power using a light microscope?
Let NA = 1.4.
The resolving power depends on the colour (or wavelength) of light. If looking at green light (the colour eyes are most sensitive to), ? = 500nm, thus:
r = 0.61 x 500nm / 1.4 = 218nm
If using blue light (which has the smallest wavelength), resolution is:
r = 0.61 x 400nm / 1.4 = 174nm
If using red light (which has the largest wavelength), resolution is:
r = 0.61 x 700nm / 1.4 = 305nm
So, the range of "best" resolution is about 200nm to 300nm

Notes:

* the wavelength of visible light ranges from 400-700nm
* A distance of 1/3 ? between two objects will have no resolution.

* An electron microscope has a better resolution since electrons have a shorter wavelength than light. In fact, the highest resolving power of an electron microscope is about 0.1nm. This is about 1000 times better than that of a light microscope!
* An increase in magnification does not help you see finer details when resolution is at its max since there is no detail to be seen. In fact, brightness decreases when the magnification increases so a greater magnification may hinder the resolution of the image.

Final Comments

Please keep in mind that I have frequently used the assumption that the object is self illuminating. A light source is needed to reflect light off the object in order for it to be seen through a microscope.
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