Interference effects for continuous sources:

i) Light bends around corners.

ii) “Shadows” fill in

iii) “Parallel” beams always spread

iv) Limits of resolution of microscopes and telescopes

Diffraction Ch 38Diffraction Ch 38

Fraunhofer Diffraction: (easy math)Source, screen “ at infinity”, ~parallel rays

eg. Laser & narrow slit.

Fresnel Diffraction: (complicated math)

Source distance, object size and screen distance all comparable.

eg.

Source Steel Ball Screen

ShadowPoisson

Spot

Single Slit, width = Single Slit, width = aa

a

θ 54321

Use Huygens’s idea:Treat the slit as a large number of point sources.

x = a sin θ is path difference between waves at two edges

(Fraunhoffer diffraction)

Divide slit in half and add up rays in pairs:

Ray (1) travels farther than ray (3): when it is ½ cycle behind -> Cancel

Ray (2) travels farther than ray (4): when it is ½ cycle behind -> Cancel

Ray (3) travels farther than ray (5): when it is ½ cycle behind -> Cancel

1) ANY two waves that originate at points a/2 away are out of phase by 180o for some angle θ, and cancel.

2) waves from upper half interfere destructively with waves from the upper half

a θ

54321

sin2

a a

a

sin

2sin

2

By dividing the slit into four parts rather than two, we can

similarly obtain:

a

2sin

By dividing the slit into six parts rather than two, we cansimilarly obtain:

a

3sin

Therefore, the general condition for destructiveinterference is:

...3,2,1,sin ma

m

where m is the order number, for dark fringes

m = 3

m = 2

m = 1

m = -1

m = -2

m = -3

m=1.5

m=-1.5

θ ...3,2,1,sin ma

m

The equation:

Gives us the positions(θ) of the MINIMA !!!

The maxima are approxhalf way between theminima.

Quiz:

What would the pattern look like if the width of the slit was increased?

A) The minima and maxima would be spaced very far apart

B) No change, the pattern would look the same

C) The minima and maxima would be spaced very closely together

Example 1: Light of λ=580nm is incident on a slit of width 0.30 mm. The observing screen is placed 2.0 m from the slit.

a) Find the positions of the first dark fringesb) Find the width of the central bright fringe

Example 2: Light of λ=580nm is incident on a slit of width 0.30 mm. The observing screen is placed 2.0 m from the slit.

a) Determine the width of the first order bright fringe.b) How does it compare to the width of the central max?

= 600 nm; central peak is 6cm wide on a screen 3m away. How wide is the slit?

slit

Example Example 33

L=3.0 mθ1

θ1

3 cm =y1

3 cm

Intensity distribution for a single slit of width a

2

max 2

sin ( / 2)( / 2)2

= sin

I I

and a

Intensity

What are the values of m and ?

a sin θ = m λ =m2

m = ±1, ±2, ±3, …(but not m = 0)

θ

54321

siny

∆y

sin

2

y

sin

2y

The intensity is a resultant of all the incremental electric fields from each part of the slit. These are out of phase by ∆β:

sin

2sin

2ayNN

(Since asinθ=mλ)

Example 4:

Show that:

a) the Minima happen when: a sinθ = m λ, m =±1, ±2…

b) to what angles of β/2 does this correspond?

Intensity

β/2

I1I2 I2

I1

π-π-2π 2π-3π 3π

Example 5: Find the intensities of the first two secondary maxima for the single slit Fraunhofer diffraction pattern:

(Hint: assume that this occurs midway between the minima.)

Notes:

1) Central maximum is twice as wide, and much brighter (~ 75% of light)

2) Secondary maxima get fainter as we move to higher orders m

3) Minima are at

4) Secondary maxima are approximately halfway between the minima. That is:

sin , 02

a m mm

1

2sin( ) ( ) for 1,2....

2

a m m

m