# Presentation: 686255

## More interference

## Interference

- Two waves
- difference in path length!
- Destructive
- Constructive
- Ways to get path length differences
- double slit
- interferometer (spit the beam)
- thin film
- Wedges
- Small apertures
- slit
- aperture...

- sizes of apertures of order of a wavelength

## So what waves?

- Sound – air waves
- Waves in ocean – water
- Music – waves in a string
- So for light
- What waves?

- Michelson Interferometer
- Albert Michelson
- 1869 US Navel Academy
- 1878 Measure speed of light
- 1882 Michelson-Morley
- Experiment at Case Tech in Cleveland: measure speed of earth’s motion in “Ether”.  Null
- result led to Einstein’s Theory of Special Relativity
- 1907 First American to win Noble Prize (Physics)

- Ether ?

**Fig 37-22, p.1194**

- Ether Wind???

**Notes:**
Active Figure 37.22 Diagram of the Michelson interferometer. A single ray of light is split into two rays by mirror M0, which is called a beam splitter. The path difference between the two rays is varied with the adjustable mirror M1. As M1 is moved, an interference pattern changes in the field of view.

## Moving through the ether

- _[shockwave applet – from publisher](file:///../AF_3722/AF_1609.html)_

## NO ETHER !

## ymmL/d   Bright Fringes

- Path Difference for Two
- Parallel Rays: *** =r**1** - r**2** = d sin***

- Constructive Interference
*d sin**** = ****m*  (All *m*, +/-)
- Destructive Interference
*d sin**** = (m+1/2)***

- L>>d   sin ~  = ym/L = m/d
- ymmL/d   Bright Fringes
- ym = (m+1/2) L/d   Dark Fringes

- m = “order” of fringe
- m = 0  zeroth order, maximum, central bright fringe
- m = +/- 1  first order maxima

## phase difference

- 1

- 1

- =

- 2

- 

- =

- 2

## Intensity Pattern of Fringes

- 2 in phase ~ 

##        v=fλ   

- n=1 in a vacuum
-  n ≥ 1 _***always***_
-  the larger n is the slower light moves

-  If  were different above and below the force on a charge would change across the interface if ***cos(**********1******t) ≠ cos(**********2******t)***
-  If *******F **********x ≠ 0*** work is done on the system
-  Energy would be deposited ...and heat up...which does not happen
-  So*** **********1******=**********2****** ***and*** **********1****** ≠**********2 ***or,

-        v=fλ
- Does f change or λ?

## Phase Changes with Reflection

*t*

-  Thin Film with n1 = n3 = 1 (air)
-  Near ***normal*** reflection ( ~ 0)
-   phase change at front surface only

- light: phase flip (by ) if n1<n2

- flip

## 6. When a light ray in a less dense medium is reflected at the surface of a denser medium, the reflected ray has a phase difference of ____ radians.
a. π/2
b. 0
c. π
d. π/3
e. π/4

- a. π/2
- b. 0
- c. π
- d. π/3
- e. π/4

## Example: Suspended Thin Soap Film

- Use white light with = 400-690 nm
- Use n = 1.33 (water)
- Use t = 320 nm
- Find  when reflected light is brightest
- Use 2nt = (m + 1/2) --->

- Note: If t < 0.1, for very
- thin films, reflection always
- dark, independent of  and
- light intensity.  This can
- occur at the very top of a
- vertical soap film.

## Reflective Coatings on Lens, Glasses & Solar Cells: Reduces Reflection-Maximizes Transmission

*t*

-  Let  = 550 nm
-  ~0 degrees
-  Want ***r******1*** and ***r******2***
-  out of phase
- for destructive
- interference and
- no reflection
- (and maximum
- transmission)

- flip

- flip

- Use 2n2t = (m+1/2)
- m = 0 for minimum thickness
- t = l/(4n2) = (550 nm)/4(1.38) = 99.6 nm

## 7. When a light ray in a denser medium is reflected at the surface of a less dense medium, the reflected ray has a phase difference of ____ radians.
a. 0
b. π/2
c. π/3
d. π/4
e. π
ANS: a

- a. 0
- b. π/2
- c. π/3
- d. π/4
- e. π
- ANS: a

## Interference from Tapered Thin Wedge

- Red light:  = 632.8 nm

- Some of light is reflected back from top and bottom surfaces of air wedge which acts like a thin film of air.
- Six dark fringes and five bright fringes are seen.
- Find L = LR - LL
-  phase at n2-n3 surface
- For dark fringes: 2L = mn2
- 2LL = mL(/n2) and
- 2LR = (mL + 5)(/n2)

- L=LR - LL = (5)/(2n2)
-      = (5)(632.8 x 10-9)/(2)(1.00)
-      = 1.58 x 10-6 m

## Newton’s Rings

- Central Dark Spot (air gap).
-  First observed by Robert Hooke.
-  Measured by Newton and
-  explained by Young.
-  Use to measure the radius of curvature of lens by counting fringes.

## Slit as a source of wavelets

## Single Slit Diffraction: look for minima

- Rays #1 and #2 interfere destructively
- ( out of phase) if

- Then pairs of rays *a/2* apart from entire gap will be out of phase.

- Central Maximum at  = 0 so no *m = 0*

- Condition for Minima

- divide by 4

- divide by 6

-  etc

## Single Slit Diffraction Intensity 

- Central maximum is twice size of spacing between mimima
- i.e between m=-1 and m=+1 so m=2

- do an integration over the slit and get

- note: β not small!

## Example 1

- Find the intensity of the first three secondary maxima in a single-slit diffraction pattern relative to the intensity of the central maximum.  Use  = (m+1/2) for the approximate angular location of the secondary maxima.

## So isn’t aren’t there two things going on in a double slit experiment?
Double slit
Single slit

**Fig 37-1a, p.1178**

- So isn’t aren’t there two things going on in a double slit experiment?
- Double slit
- Single slit

-  slit width a

- d slit separation

**Notes:**
Figure 37.1 (a) If light waves did not spread out after passing through the slits, no interference would occur.

## Diffraction from Double Slit

- Double slit

- Single slit

-  realistic Double slit

**Fig 38-11, p.1213**

**Notes:**
Active Figure 38.11 The combined effects of two-slit and single-slit interference. This is the pattern produced when 650-nm light waves pass through two 3.0- mm slits that are 18 mm apart. Notice how the diffraction pattern acts as an “envelope” and controls the intensity of the regularly spaced interference maxima.

**Fig 38-14, p.1217**

- eye
- lens

-  eye
- retina

**Notes:**
Figure 38.14 (Example 38.4) Two point sources separated by a distance d as observed by the eye.

## Diffraction by a Circular Aperture

- Bright central maximum is called “Airy Disk” (Sir George Airy)
- Resolution Limit: Rayleigh’s Criterion (Lord Rayleigh, 1899, Asked “why is the sky blue”, Noble Prize in 1904, discovered Argon)---Center of 2nd Airy Disk at 1st minimum of 1st Airy Disk
- a=diameter

- 1

- 1

- remember for small
- sin

## Fig 38-13, p.1215

**Fig 38-13, p.1215**

**Notes:**
Figure 38.13 Individual diffraction patterns of two point sources (solid curves) and the resultant patterns (dashed curves) for various angular separations of the sources. In each case, the dashed curve is the sum of the two solid curves. (a) The sources are far apart, and the patterns are well resolved. (b) The sources are closer together such that the angular separation just satisfies Rayleigh’s criterion, and the patterns are just resolved. (c) The sources are so close together that the patterns are not resolved.

## Example 2

- Limiting Resolution of Human Eye
- sin=1.22λ/a
-  = 550 nm    *a *= 4 x 10-3 m  ( 4 mm)
- = (1.22x550x10-9)/(4x10-3) = 1.7x10-4 radians (30 arcsec)
- Telescope: *a *= 8 m
- = (1.22x550x10-9)/(8) = 8.4x10-8 radians (0.017 arcsec)

## 8. A thin film of thickness t and index of refraction n is surrounded by air on both sides. Constructive interference will occur in light of wavelength  in air upon reflection from the film when
a. .
b. .
c. .
d. .
e. .
ANS: b

- a. .
- b. .
- c. .
- d. .
- e. .
- ANS: b

## Poison’s (Arago’s) Bright Spot

-  18th Century Science Controversy in France
-  Newton’s Corpuscular Theory of Light was the Most Popular
-  Augustin Fresnel developed wave theory to explain diffraction
-  Simeon Poison used wave theory to predict *bright spot* at center
-   of shadow of dark spot--this was hard to see so he thought this proved
-   that the wave theory was wrong.
-  In 1818, Dominique Arago found *bright spot*.

**Object**

**Shadow w/ **
**Diffraction Pattern**

- Bright Spot

- Region outside
- Geometric Shadow
- has Fringes-Edge
- not sharp

**Fig 38-3, p.1207**

**Notes:**
Figure 38.3 Diffraction pattern created by the illumination of a penny, with the penny positioned midway between screen and light source. Note the bright spot at the center.

## Diffraction Gratings

- Diffraction from Double Slit

## Diffraction Gratings

- Intensity of maxima: *I ~ N**2      **N = # lines in grating*
- Maxima at *dsin**** = m***  all m     *d* is grating spacing
- Central maximum at *m = 0*, satisfied for all 
- Width of maxima: *** ~ 1/N*
- Grating Dispersion: *D = **** = m/dcos***
- Grating Resolving Power:  *R **** = Nm*

## Example

- Grating: N = 1.26 x 104 lines in width w = 25.4 mm
- Sodium Lamp w/ Doublet:   1 589.00 nm, 2 = 589.59 nm
- Consider 1st order maximum, m = 1
- Line spacing *d = w/N* = (25.5 x 10-3 m)/1.26 x 104 = 2016 nm
- Using d sin = m      = sin-1(/d) = 17.0o
- From Dispersion D =  (m/d sin)
-   = (589.59 - 589.00)(1)/(2016 cos 17o)
-  = 3.06 x 10-4 radians = 0.0175o
- Resolving Power  R:  R =  = Nm = (589.00 nm)/(0.59 nm)
- Therefore N = 999 is sufficient to resolve doublet at 1st order.

## Holography

- Laser Beam

- Mirror

- Mirror

- Beam Spitter

- Diffusing Lens

- Film records interference
- patterns, both phase & intensity.
- Used to reproduce 3-D images

- Plane waves

- Spherical Waves

- Interference pattern on film acts as diffraction grating

- Real Image

- Virtual Image

- Viewing

## 9. A thin film of thickness t and index of refraction n is surrounded by air on both sides. Destructive interference will occur in light of wavelength  in air upon reflection from the film when
a. .
b. .
c. .
d. .
e. .
ANS: e

- a. .
- b. .
- c. .
- d. .
- e. .
- ANS: e