# Calculate Resolving Power of Microscope w/ 1.30cm Aperture

• six789
In summary, the conversation discusses the problem of calculating the resolving power of a microscope with a 1.30cm aperture using 540nm light. The conversation provides a solution hint, which involves using the formula θ=1.22(λ)/D and accounting for the difference between the wavelength in air and in lens glass. The conversation also includes a link for further explanation and clarification on the derivation of the formula. The conversation ends with the solution being confirmed as correct and a question about converting the answer from radians to degrees.
six789
Here is the problem...
Calculate the resolving power of a microscope with a 1.30cm aperture using 540nm light. The index of refraction of the lens slows the light inside the glass to 1.98x10^8m/s.

i just know the formula tetha=1.22(lambda)/D, but i don't know how to derive the formula since i don't know any eqaution related to this matter.. Help me please...

Why do you feel you need to rederive the equation for resolution in order to solve the problem?

six789 said:
Here is the problem...
Calculate the resolving power of a microscope with a 1.30cm aperture using 540nm light. The index of refraction of the lens slows the light inside the glass to 1.98x10^8m/s.

i just know the formula tetha=1.22(lambda)/D, but i don't know how to derive the formula since i don't know any eqaution related to this matter.. Help me please...
SOLUTION HINTS:
This is a case of Airy Diffraction from a circular aperture (a special case of Fraunhofer Diffraction). The angular resolving power is (approx) given by:

$$1: \ \ \ \ \color{black} \theta_{resolve} \ \approx \ \frac{(1.22)\lambda}{(Aperture \ Diameter)}$$

The difference between λ in air and that in lens glass must also be accounted for:

$$2: \ \ \ \ \frac{c}{\color{blue}v_{lens}} \ = \ \frac{\lambda_{air}}{\color{red}\lambda_{lens}}$$

We are given that:
{Aperture Diameter} = (1.30 cm) = (1.3e(-2) meters)
{Wavelength in Air} = λair = (540 nm) = (5.4e(-7) meters)
{Light Speed in Lens} = vlens = (1.98x10^8 m/s)

Solving for lens" in Eq #2, we get:
{Wavelength in Lens} = λlens =
= (5.4e(-7) meters)*(1.98x10^8 m/s)/(3.0x10^8 m/s) =
= (3.564e(-7) meters) ::: <---- lens" calculated using Eq #2 with other given values

With Eq #1 above, calculate θresolve using {λ = λlens} and the other given parameter values.

Derivation and discussion of Airy Diffraction from a circular aperture (a special case of Fraunhofer Diffraction) can be found here:
http://scienceworld.wolfram.com/physics/FraunhoferDiffractionCircularAperture.html
A brief overview discussion can be found here:
http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/cirapp.html

~~

Last edited:
space tiger, what i mean is how can i derive another formula, coz

tetha=1.22(lambda)/D

this is the only formula i know for now... since there is speed, how can i apply it to the formula... can u derive me some formula and can u explain it step by step...

six789 said:
space tiger, what i mean is how can i derive another formula, coz

tetha=1.22(lambda)/D

this is the only formula i know for now... since there is speed, how can i apply it to the formula... can u derive me some formula and can u explain it step by step...
When light travels thru a lens, its speed and wavelength change and are different from those in air. (You may want to review Index of Refraction concepts for more info.) Because the angular resolving power "θresolv" depends on "λ", the light's lens" while it travels thru the lens must be used.

All necessary formulas are provided in Msg #3 of this Thread, including Eq #2 which relates light speed "vlens" (<--- value given in problem) and lens" for light traveling thru the lens. Study the indicated formulas and calculations, and perform the last computation indicated in RED (with Eq #1) using the calculated value for lens" together with the other parameter values listed.

~~

ok xanthym, i get it now... all the whole thing... i can grasp what u mean, but where did the 2nd formula came from in 3rd messsage?? is it a relationship in what?

ok, i seem to understand it, but why is my answer incorrect to the book? my answer is 3.344676923x10^-5 whereas in the book, it is 0.00192 degrees?

Your answer is correct. Convert it to degrees and you'll find it's the same.

u mean to say it is in radians? how can i convert it to degrees? is it divide by 180 or times 180?

ok xanthym, i get the whole concept now... thanks for the effort.. i now understand.. =)

## What is the resolving power of a microscope?

The resolving power of a microscope refers to its ability to distinguish between two closely spaced objects as separate entities. In other words, it is the ability to see fine details and structures in an image.

## How is the resolving power of a microscope calculated?

The resolving power of a microscope can be calculated using the formula: R = 0.61λ/NA, where R is the resolving power, λ is the wavelength of light used, and NA is the numerical aperture of the objective lens.

## What is the numerical aperture of a microscope?

The numerical aperture (NA) of a microscope is a measure of the lens' ability to gather and focus light. It is calculated as NA = n sin α, where n is the refractive index of the medium between the objective lens and the specimen, and α is the half-angle of the cone of light entering the lens.

## What is the relationship between numerical aperture and resolving power?

The resolving power of a microscope is directly proportional to the numerical aperture of the objective lens. This means that a higher numerical aperture will result in a higher resolving power, allowing for better visualization of fine details in an image.

## How does the aperture size affect the resolving power of a microscope?

The aperture size of a microscope is a crucial factor in determining its resolving power. A larger aperture, such as 1.30cm, will result in a higher numerical aperture and thus a higher resolving power. This is because a larger aperture allows for a wider cone of light to enter the lens, resulting in better light collection and increased resolution.

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