Diffraction

Physics 202Professor Lee

CarknerLecture 26

PAL #25 Interference Applications

Wavelength of laser D = d = 0.25 mm = 0.00025 m y = 1.5 cm = 0.015 m (between any 2

maxima) = yd/D = (0.015)(2.5X10-4)/(5.5) = 682 nm

Is this reasonable?

Diffraction When light passes though a small aperture it

spreads out

This effect also occurs when light passes by an obstacle

The pattern consists of minima and maxima of decreasing intensity as you move away from the center

Diffraction and Optics

Real point sources of light always experience diffraction and so we never see true point images

Diffraction limits even the best optics

Degree of diffraction (and blurriness) depends on aperture size and wavelength

Diffraction and Interference

Light rays from different parts of the same aperture can also produce interference

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

Single Slit Diffraction

Location of the Minima Since:

L /d = sin

How far apart can a pair of rays get?

For the first minima L must equal /2:

a sin =

Diffraction Patterns Since a sin = for m=1, we can say for the more

general case:a sin = m(min)

The maxima are located halfway between the minima

Since waves from the top and bottom half cancel

Smaller slit means more flaring

Rays from Slit

Intensity

Intensity of maxima decrease with increasing

We will use which is half the phase difference between the top and bottom of the slit

= ½ = (a/) sin I = Im (sin /)2

Diffraction and Interference

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 = 1.5, 2.5, 3.5 … maxima

Diffraction and Circular Apertures

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

For m = 1

The minima and maxima appear as concentric circles

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

Resolution and Circular Aperture

Rayleigh’s Criterion

This will be true if their angular separation satisfies the diffraction formula

R = 1.22 /d

This is called Rayleigh’s criterion