# Single lens creating a diffraction pattern?

jisbon
Homework Statement:
A single-lens generates a diffraction pattern of dark and bright fringes on a screen
located 2 m away from the lens. If the first and the second dark fringes are
separated by 0.1 mm distance, determine the lens diameter. Assume that the light
source operates at 633 nm of wavelength. What is the minimum distance that the
lens can resolve on the screen?
Relevant Equations:
##d\sin \theta =m\lambda##
Hi all,

So far all the problems I dealt with is dealing with double slits when working with dark and bright fringes. In this case, what should I do in regards with a lens? Also, what does the question mean when it asks for the minimum distance that the lens can resolve on the screen? Does it mean the minimum distance between the lens and the screen to form the fringes?

Assuming I can treat them as a double slit,

##d\sin \theta =(m+1/2)\lambda##
First dark fringe m= 0, 2nd dark fringe m=1,

From there I can find and compare the distance etc.

So the question is can I treat it as double-slit? If so why is it? Thanks

Homework Helper
Gold Member
Homework Statement:: A single-lens generates a diffraction pattern of dark and bright fringes on a screen
located 2 m away from the lens. If the first and the second dark fringes are
separated by 0.1 mm distance, determine the lens diameter. Assume that the light
source operates at 633 nm of wavelength. What is the minimum distance that the
lens can resolve on the screen?
Homework Equations:: ##d\sin \theta =m\lambda##

Hi all,

So far all the problems I dealt with is dealing with double slits when working with dark and bright fringes. In this case, what should I do in regards with a lens? Also, what does the question mean when it asks for the minimum distance that the lens can resolve on the screen? Does it mean the minimum distance between the lens and the screen to form the fringes?

Assuming I can treat them as a double slit,

##d\sin \theta =(m+1/2)\lambda##
First dark fringe m= 0, 2nd dark fringe m=1,

From there I can find and compare the distance etc.

So the question is can I treat it as double-slit? If so why is it? Thanks
You shouldn't compare it to a "double" slit pattern. Double slits produce an "interference" pattern superimposed on the diffraction pattern of a given, single slit. I think you might be confusing interference with diffraction.

That said, you have the right equation. Comparing this to a single slit diffraction pattern is acceptable. [Edit: Well, your first equation is correct. However, realize that in circular diffraction, $m$ is not an integer, nor is it evenly spaced between the first and second minimums. $m = 1.22$ for the first minimum and $m = 2.233$ for the second. You'll need to do a little more research to compare circular diffraction to single slit diffraction.]

Assuming the lens is circular in shape, then the diffraction pattern will produce what's called an "Airy disk." And yes, the rings on the Airy disk follow the same formulae as single slit diffraction. [Edit: see my edit above regarding some differences.]

[Edit: Do a google search or look up the Wikipedia article on "Airy disk" for some interesting and useful information.]

Last edited:
Homework Helper
Gold Member
Also, what does the question mean when it asks for the minimum distance that the lens can resolve on the screen? Does it mean the minimum distance between the lens and the screen to form the fringes?
Oh, and regarding your second question:

The "resolving power" of lens is the minimum angular distance between two points a lens can distinguish. It is a function of lens' diameter and the wavelength of light used. You'll need another formula for this (it wasn't listed in your "Homework equations" section). It's similar to the diffraction formula, but not identical.

You'll need to convert this angular distance to linear distance on the screen. So no, the question is not asking for the minimum distance between the lens and the screen. The question is asking for the minimum distance between two points on the screen that the lens is able to resolve.

It is generally accepted that we can resolve two objects in an image if the peak of the central maximum of one of the sources lies on or outside of the the first minima of the diffraction pattern of the other source; the critical distance is known as the Rayleigh criterion.

The first minimum of an Airy disk occurs at ##\theta = \frac{1.22\lambda}{d}##. And if ##\theta## is small, you can use ##\tan{\theta} \approx \theta##.

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Cutter Ketch
It is interesting that this question gives the distance between the first two minima. More usually problems will give the distance from the center to the first minimum. That is the 1.22 ##\frac {\lambda} d## others have mentioned. The reason they did that is because the Airy pattern produced by diffraction through a round aperture is significantly different than the one dimensional pattern produced by a single slit. For example, if you compared the angle to the first minimum they differ by 1.22. However it just so happens that the angular distance between the first and second minimum for a round aperture of diameter d and a slit of width d are the same to within about 1%.

• etotheipi
jisbon
I was taught this formula, but as far as I remembered this formula was only used for resolution purposes. In this case, how do I determine that for the first minimum and m=2.233m=2.233 for the second minimum? I can't really understand this part?

Cutter Ketch
I was taught this formula, but as far as I remembered this formula was only used for resolution purposes. In this case, how do I determine that for the first minimum and m=2.233m=2.233 for the second minimum? I can't really understand this part?

In the formula you give for the location of the minima in single slit diffraction m is an integer. That formula is correct, and that m is exactly an integer.

The problem is that a round aperture doesn’t work like that. An exact expression for the diffraction pattern can be written, but the answer contains ##J_1##, the first order Bessel function of the first kind.

I(##\theta##) = ##I_0 {\left[ \frac {2 J_1(x)} x \right]}^2##

You can solve the equation to find the angular location of the zeros. However the zeros of the function aren’t related by integers. They aren’t even equally spaced. We probably shouldn’t equate them logically with the integer m in the single slit case. You can look up the definition of the Bessel function and find the zeros, or you can just look them up. See https://en.m.wikipedia.org/wiki/Airy_disk

• BvU and collinsmark