Calculate Resolving Power of Microscope w/ 1.30cm Aperture

[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...

 

[SpaceTiger]

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

 

[xanthym]

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} = λ_air = (540 nm) = (5.4e(-7) meters)
{Light Speed in Lens} = v_lens = (1.98x10^8 m/s)

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

With Eq #1 above, calculate θ_resolve using {λ = λ_lens} 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.)

~~

 

[six789]

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...

 

[xanthym]

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 "θ_resolv" depends on "λ", the light's "λ_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 "v_lens" (<--- value given in problem) and "λ_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 "λ_lens" together with the other parameter values listed.


~~

 

[six789]

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?

 

[six789]

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?

 

[Data]

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

 

[xanthym]

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?

Derivation and discussion given here:
http://www.rpi.edu/dept/phys/Dept2/APPhys1/optics/optics/node7.html


~~

 

[six789]

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

 

[six789]

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