Circular aperture

  • Thread starter six789
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  • #1
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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 dont know how to derive the formula since i dont know any eqaution related to this matter.. Help me please...
 

Answers and Replies

  • #2
SpaceTiger
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Why do you feel you need to rederive the equation for resolution in order to solve the problem?
 
  • #3
xanthym
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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 dont know how to derive the formula since i dont 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:

[tex] 1: \ \ \ \ \color{black} \theta_{resolve} \ \approx \ \frac{(1.22)\lambda}{(Aperture \ Diameter)} [/tex]

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

[tex] 2: \ \ \ \ \frac{c}{\color{blue}v_{lens}} \ = \ \frac{\lambda_{air}}{\color{red}\lambda_{lens}} [/tex]

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
(For this latter ref, follow the indicated links for additional info.)

~~
 
Last edited:
  • #4
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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...
 
  • #5
xanthym
Science Advisor
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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.


~~
 
  • #6
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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?
 
  • #7
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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?
 
  • #8
998
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Your answer is correct. Convert it to degrees and you'll find it's the same.
 
  • #10
127
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u mean to say it is in radians? how can i convert it to degrees? is it divide by 180 or times 180?
 
  • #11
127
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ok xanthym, i get the whole concept now... thanks for the effort.. i now understand.. =)
 

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