Resolving Power of a Diffraction Grating - Notes | PHYS 262, Study notes of Physics

Material Type: Notes; Class: Recitation for Lecture 001; Subject: Physics; University: George Mason University; Term: Unknown 1989;

Typology: Study notes

Pre 2010

Uploaded on 02/10/2009

koofers-user-q5r-1
koofers-user-q5r-1 🇺🇸

9 documents

1 / 12

Toggle sidebar

This page cannot be seen from the preview

Don't miss anything!

bg1
Resolving Power of a Diffraction
Grating
From our phasor discussion in relation to the multiple minima between adjacent
maxima in a multi-slit diffraction pattern, the phase difference
between
adjacent minima is given by 2
/N, where Nis the number of slits.
1st adjacent min
2
N

pf3
pf4
pf5
pf8
pf9
pfa

Partial preview of the text

Download Resolving Power of a Diffraction Grating - Notes | PHYS 262 and more Study notes Physics in PDF only on Docsity!

Resolving Power of a DiffractionGratingFrom our phasor discussion in relation to the multiple minima between adjacentmaxima in a multi-slit diffraction pattern, the phase difference

^ between

adjacent minima is given by 2

/ N , where^ N^ is the number of slits.^2   ^ N st 1 adjacent min^ 

Resolving Power of a DiffractionGrating Consider infinitesimal changes, we have the following relationship,

2 cos d^ ^    2 2 cos

cos d^ orN

Nd ^ ^ 

 

 ^ ^

^  Thus, for the small change in phase

, the angular position will change by,

sin^2 sin^2 N^ ^ ^  d^ 2

dor ^ 

^  ^ 

 ^

 We also know how phase difference are related to the angular location

, This gives the ½ widthof the peak at order^ m.

Resolving Power of a DiffractionGratingSo, we equate the two boxed quantities,

R Nm  ^  Canceling the common factor

m  cos cos Nd d d cos , we finally arrived at the desired relation,

    So, the resolving power of a diffraction grating increases by• using a grading with a larger number

N^ of lines

  • measuring the spectra line at a higher order

m

Resolving Power (again) for CircularApertures^ Because of

diffraction , light spreads out after passing thru smallapertures ^ this posses^ resolution limits

to commonly used optical instruments, such as microscopes and telescopes.

image will not be sharp

Resolving Power for CircularAperturesThe overlap of the two diffraction pattern might prevent one from discerningthe two sources of light.A workable criterion is called the

Rayleigh’s Criterion

which is similar in

spirit to our discussion for the resolving power for the diffraction grating:The two diffraction pattern will be resolvable if thecentral max from one pattern is at least as far as thest^1

min of the other image. For circular aperture with diameter

D , the location of the its 1

st^ order diffraction

minimum is:

sin 1.22 (^)   1 D (“1.22” is a geometric factor)

Resolving Power for CircularAperturesThe^ Limit of Resolution

for a circular aperture is defined as the smallest angular separation between two light sources that can be resolved accordingto the^ Rayleigh’s Criterion

and it is given by:sin^ 1.22^ min

   D

Example 36.6: Resolving Power of aCamera Lens^ Given : f=50 mmf-number of f/2object distance 9.0mwavelength = 500nm

What is the minimum distance between twopoints on the object that one can resolve? f-number = f/D

^ D = f/f-number = 50mm/2 =25 mm Rayleigh’s Criterion gives:

9 5 min^ min^

500 103 sin^ 1.^

1.^ m^ 2.4^10 rad 25 10 D^

m  

  ^ ^

^

^ 

Example 36.6: Resolving Power of aCamera LensFor a simple lens, we know that the angular separation of two points on theobject is given by,

' y y   ' s s y^ y ^ separation of object points y’ ^ separation of the corresponding image points s ^ object distance s’ ^ image distance

y’ s

s’