Lecture 16 Diffraction Ch. 36

• Topics– Newtons Rings– Diffraction and the wave theory– Single slit diffraction– Intensity of single slit diffraction– Double slit diffraction– Diffraction grating– Dispersion and resolving power

• Demos– laser pointer internal reflected in water flow– laser pointer interference from many reflections– Youngs double slit interference with narrow slits– Newtons rings– Laser diffracted from bead– Laser defracted from single slit– Laser diffracted from single slits of different widths– Now show diffraction and interference together using two slits whose width is

a few wavelengths– Measure diameter of hair using laser .– Show diffraction using gratings with different number of linesElmoPolling

Newton’s rings are similar, but with curved glass:

Why different colors? for any given difference in path length,the condition ΔL =n (m+1/2)λ might be satisfied for somewavelength but not for some other. A given color might ormight not be present in the visible image.

Pass around Newtons Rings

Chap. 35 Problem 75The figure shows a lens with radius of curvature R lying on a flat glass plate andilluminated from above by light with wavelength λ. The radius of curvature R of the lens is5.0 m and the lens diameter is 20 mm.

a) How many bright rings are produced? Assume that λ = 589 nm.(b) How many bright rings would be produced if the arrangment were immersed in water (n= 1.33)?

Diffraction

On a much smaller scale, when light waves passnear a barrier, they tend to bend around thatbarrier and spread at oblique angles.

This phenomenon is known as diffraction of the light,and occurs when a light wave passes very close tothe edge of an object or through a tiny opening,such as a slit or aperture.

Lets look at some experiments to see what theeffect looks like

Diffraction from a straight edge

Interference pattern of light and dark bandsaround the edge of the object.

Diffraction is often explained in terms of theHuygens principle, which states that each pointon a wavefront can be considered as a source ofa new wave. This leads to phasors.

All points on a wavefront serve aspoint sources of spherical secondarywavelets. After a time t, the newposition of the wavefront will bethat of a surface tangent to thesesecondary wavefronts

He-Ne laser diffracted from a straight edge

Calculated using the Cornu Spiral from phasors

However, Dominique Arago, another member of the judging committee,almost immediately verified the spot experimentally. Fresnel won thecompetition, and, although it may be more appropriate to call it "theSpot of Arago," the spot goes down in history with the name "Poisson'sbright spot" like a curse

In 1818, Augustin Fresnel submitted a paper onthe theory of diffraction for a competitionsponsored by the French Academy. S.D. Poisson,a member of the judging committee for thecompetition, was very critical of the wavetheory of light. Using Fresnel's theory, Poissondeduced the seemingly absurd prediction that abright spot should appear behind a circularobstruction, a prediction he felt was the lastnail in the coffin for Fresnel's theory.

The Fresnel bright spotDiffraction from a opaque disk

Diffraction from a human hair

Experiment to measure diameter of teachers hairusing first diffraction minimum.

asin! = " for the first diffraction minimum

a ="

sin!

sin! # tan! =y

1

L

a ="Ly

1

a = 560 nm L

y1

$

%&'

()

Location of firstdiffraction minimum

L

screen

hair y1

a

Diffraction by a Single Slit - Locate the First Minima

D >> a, rays are parallel

Divide the screen into two zones of width a/2

Find the first minima above the midpoint bypairing up rays from the top point of the topzone and the top point of the bottom zone

Rays are in phase at the slit but must be outof phase by by λ/2 at screen

The path length difference = a/2 sin θ

Repeat for other pairs of rays in the upperzone and lower zone: a/2 sin θ = λ/2

a sin θ = λ first minima

If we narrow the slit the angle must get bigger - more flaring - - what happens when a = λ?

Diffraction by a Single Slit -Subsequent MinimaRepeat the process for paired rays (4) fromcorresponding points from each of the zones

Rays are in phase at the slit but must be outof phase by by λ/2 at screen

second minimum

Dark fringes - minima

Intensity in Single Slit Diffraction - Deriveusing phasors - see text

where

α is just a convenient connection between the angle θ that locates apoint on the screen and the light intensity I(θ)

I = Im occurs at the central maximum (θ = 0) and ϕ is the phasedifference (in radians) between the top and bottom rays from the slitof width a

Studying the expression above reveals that the intensity minima occurat

Plug this into the expression for α above and we find

alpha

Divide aperture up into 18 zones. Each zone is asource of waves Hyugens principle

See Fig 36-6 Example of phasor diagrams

sin θ

Exact solution for diffraction maxima

�

I(!) = I0

sin2"

" 2

#

$ %

&

' ( = I0(sin

2")(")2)

dI

d!= 0

dI

d!= I

02sin! cos! !

"2" I

02sin

2! !

"3= 0

Has two solutions

To find maxima of a function, take derivative and set equal to 0

1) sin! = 0

2) cos! = sin! /!

or ! = tan!

! ="a

#sin$

Minima ! = m" m=1, 2, 3..

Maxima

I02

sin!

!2

(cos! "sin!

!) = 0

Solutions are at the intersections y= α and y=tan α

1st side maximum

Central Maximum

α=4.493 rad

! = (m +1

2)" = 4.493

m = 0.93

! = (m +1

2)" = 0

m = #0.5

Exact

minima asin! = m" m=1, 2, 3..

After the central maximum the next maximum occurs about halfway between the first and second minimum.

maxima asin! " (m +1

2)# m =1, 2, 3.. Approximate

sin θ

m =1

�

sin! = m"

a= m

"

"= m =1

sin! = m"

a= m

"

5"=m

5

�

sin! = m"

a= m

"

10"=m

10

Note that as the slit gets wider the central maximum gets smaller -contrary to expectations-

Diffraction by a circular aperture (lens ...)

First minimum

Compare to the relation for a single slit

A complex analysis leads to

When light from a point sourcepasses through a small circularaperture, it does not produce abright dot as an image, butrather a diffuse circular discknown as Airy's disc surroundedby much fainter concentriccircular rings. This example ofdiffraction is of greatimportance because the eyeand many optical instrumentshave circular apertures.

ResolvabilityThe fact that a lens image is a diffraction pattern is important whentrying to resolve distant point objects

When the angular separation for two objects issuch that the central maximum of one diffractionpattern coincides with the first minimum of theother we have a condition called the Rayleigh’scriterion for resolvability

Example 15E. The two headlights of an approaching automobile are 1.4 mapart. Assume the pupil diameter is 5.0 mm and the wavelength of light is550 nm. (a) At what angular distance will the eye resolve them and (b) atwhat distance?

�

!R

= s /D

D = s /!R

D = 1.4m /134.2 "10#6

rad

D = 0.010 "106m = 10km

�

!R

=1.22" /d

�

!R

=1.22(550 "10#6mm) /5.0mm

�

!R

=134.2 "10#6rad

(a)

(b)

sD

�

!R

=1.22" /d�

!R

Diffraction and Interference by a double slitI = I (double slit interference) x I(diffraction)

�

I(!) = Imcos

2(") #

sin2$

$ 2

! ="d sin#

$ dsin#=m$ m =0,1,2.. maxima

�

! ="asin#

$ asin# = m$ m =1,2,3.. minima

How many bright interference fringes fall within the central peak of the diffraction envelope?

Sample problem 36-5

! = 405nm

d = 19.44nm

a = 4.050nm

(a) How many bright interference fringes fall within the centralpeak of the diffraction envelope?

Given

asin! = "

d sin! = m2"

m2= d / a = 19.44 / 4.050 = 4.8

We have m=0 and m=1,2,3 and 4 on bothsides of central peak.

Answer is 9

2 refers to double slit interference

Diffraction Gratings Like the double slit arrangement butwith a much greater number of slits,or rulings, sometimes as many asseveral 1000 per millimeter

Light passed through the gratingforms narrow interference fringesthat can be analyzed to determinethe wavelength

As the number of rulings increasesbeyond 2 the intensity plot changesfrom that of a double slit pattern toone with very narrow maxima (calledlines) surrounded by relatively wide darkregions

Diffraction GratingsPrincipal maxima, where all wavesinterfere constructively are found withwhat is now a familiar procedure:

each m represents a different line andthe integers are called order numbers

Rewriting the above we find

So the angle to a particular line (say m =3) depends on the wavelengthbeing used

DispersionIn order to distinguish different wavelengths that are close toeach other a diffraction grating must spread out the linesassociated with each wavelength. Dispersion is the term usedto quantify this and is defined as

Δθ is the angular separation between two lines that differ byΔλ. The larger D the larger the angular separation betweenlines of different λ.

It can be shown thatand D gets larger for higherorder (m) and smaller gratingspacing (d)

Resolving PowerTo make lines that whose wavelengths are close together (toresolve them) the line should be as narrow as possible

The resolving power is defined by

where λavg is the average of the two wavelengths studied and Δλ is thedifference between them. Large R allows two close emission lines to beresolved

It can be shown that To get a high resolving powerwe should use as many rulings.

Dispersion D and Resolving Power RThree gratings illuminated with light ofλ=589 nm, m = 1

GratingGrating NN d(nm)d(nm) θθ D(D(oo//μμm)m) RR

A 10000 2540 13.4 23.2 10000

B 20000 2540 13.4 23.2 20000

C 10000 1360 25.5 46.3 10000

Highest dispersion

Highest resolution

!"hw

=#

Nd cos"

!"hw

D =!"

!#

R =!avg

"!

Spacing between atomic layers in a crystal is the rightsize for diffracting X-rays; this is now used to determinecrystal structure

X-Ray Diffraction

Bragg’s law, θ is Bragg’s angle

Chapter 36 Problem 1

Light of wavelength 561 nm is incident on a narrow slit. Theangle between the first diffraction minimum on one side of thecentral maximum and the first minimum on the other side is0.90°. What is the width of the slit?

Chapter 36 Problem 29

Suppose that the central diffraction envelope of a double-slitdiffraction pattern contains 11 bright fringes and the firstdiffraction minima eliminate (are coincident with) brightfringes. How many bright fringes lie between the first andsecond minima of the diffraction envelope?

Chapter 36 Problem 58

An x-ray beam of a certain wavelength is incident on a NaClcrystal, at 26.0° to a certain family of reflecting planes of spacing39.8 pm. If the reflection from those planes is of the first order,what is the wavelength of the x rays?