Diffraction

Physics 202Professor Lee

CarknerLecture 24

PAL #23 Interference Applications

Wavelength of laser D = d = y = 3 mm = 0.003 m (between 0 and 1

maxima) y = mD/d = yd/D = (0.003)(0.00025)/(1.02) =

Is this reasonable? No, laser is red and red light has a

wavelength between ~ 600- 700 nm

PAL #23 Interference Applications

What color does a soap film (n=1.33) appear to be if it is 500 nm thick? We need to find the wavelength of the maxima:

= (2Ln) / (m + ½)

= [(2) (500nm) (1.33)] / (m + ½)

= 2660 nm, 887 nm, 532 nm, 380 nm …

Real soap bubbles change thickness due to turbulence and gravity and so the colors shift

Diffraction When light passes though a small aperture

it spreads out

This flaring produces an image with a bright central maximum and progressively fainter maxima at increasing angles

Geometric optics assume point images, but all real images are blurry

Diffraction and Interference

Young’s experiment is an example of light rays from two different apertures producing interference

This is called single slit diffraction

Instead of two rays from two slits, we have a continuum of rays emerging from one slit

Path Length Difference Minima (dark fringes) should occur at the point where

half of the rays are out of phase with the other half

If we assume that the distance to the screen (D) is much larger than the slit width (a) then the path difference is

where d is the distance between the origin points of the two

rays

We will pair up the rays, and find the path length for which each pair cancels out

Location of the Minima Where is the first minima? Since:

L /d = sin

How far apart can a pair of rays get?

For the first minima L must equal /2:

(a/2) sin = /2a sin =

Diffraction Patterns

a sin = m(min) Where is the location of the minima

corresponding to order m

Note that this relationship is the reverse of that for double slit interference [d sin = (m+½): min]

Since waves from the top and bottom half cancel

Intensity

Intensity of maxima decrease with increasing The intensity is proportional to the value of E2,

which in turn depends on the phase difference

= ½ = (a/) sin I = Im [(sin /]2

where Im is the maximum intensity of the pattern

Intensity Variations The intensity falls off rapidly with

linear distance y

Remember tan = y/D

The narrower the slit the broader the maximum

Remember: m = 1,2,3 … minima m = 1.5, 2.5, 3.5 … maxima

Diffraction and Circular Apertures

The location of the minima depend on the wavelength and the diameter (d) instead of slit width:

sin = 1.22 /d For m = 1

The minima and maxima appear as concentric circles

Rayleigh’s Criterion We will consider two near-by point sources to be

resolvable if the central maximum of one lies on the first minimum of the other

For small angles:R = 1.22 /d

This is called Rayleigh’s criterion Small angle is better

Short and large d give better resolution (smaller R)

Resolution Since virtually all imaging devices have

apertures, virtually all images are blurry

If you view two point sources that are very close together, you may not be able to distinguish them

Next Time

Read: 36.7-36.9 Final, Monday, Feb 13, 9-11 am

About 2/3 covers optics About 1/3 covers fluids, SMH and

waves and thermo Four equation sheets given

If the thickness of the middle layer is ½ wavelength, what kind of interference would you see?

a) Constructiveb) Destructivec) None

n=1

n=1.5

n=1.3

If the thickness of the middle layer is ½ wavelength, what kind of interference would you see?

a) Constructiveb) Destructivec) None

n=1.3

n=1.5

n=1.1

If the thickness of the middle layer is ½ wavelength, what kind of interference would you see?

a) Constructiveb) Destructivec) None

n=1.3

n=1n=1.1