Chapter 36 – Diffraction

I.Diffraction – the phenomenon of light bending around the edge of obstacles.

This effect was ignored in the previous discussions of interference, but will now be taken into account.

A.Contrast ray approach with wave approach: Take a look at light that passes the edge of a sharp object.

Geometric optics:

Physical optics:

B.Types of Diffraction: Fresnel and Fraunhofer

II.Single Slit Diffraction – analyze it by breaking up the slit into many small segments where each segment acts like a source of light waves. We will examine what happens to the intensity on the screen for different angular positions. The angle is measured from the optic axis and labeled as the angle  and the width of the slit is a.

A. = 0o

B.Another value of 

C.Another value of 

D.In general

E.What are the intensities at different angles break the slit into many little sections of width ds, and consider waves traveling toward the screen from each of the sections. If the total amplitude of the waves traveling through the slit is Em , then the amplitude of that part of the wave traveling through a width ds can be written as Em(ds/a), which represents a fraction of the total amplitude.

At point P on the screen the electric field due to top and bottomwaves is

dE = dE+ + dE- = 2Emcos  sin (kr - t)

F.Where do the maxima and minima occur? Find = 0.

1.Minima:

2.Maxima:

G.What are the intensities?

:

:

 = 

:

= 2

:

= 3:

:

III.Real Double Slits

Example: If d = 5a, then how many interference maxima are within the central maximum of the diffraction pattern?

IV.Multiple Slits

The phasor method can be used to easily determine the amplitude of the resultant wave reaching a point on the screen for any number of slits.

A.Three Slits:E1 = Emsin (t)

E2 = Em sin (t + )

E3 = Em sin (t + 2) ,

where  = (2)d sin  .

B.Draw the intensity pattern:

 = 0:I =

 = 2I =

 = 

 = (2)/3:I =

 = (4)/3:I =

C.Graphs of intensity

V.The Diffraction Grating

A.Used to separate various wavelengths of light to different angular positions. Knowing the angular position will give the wavelength.

Consider monochromatic light passing through a series of parallel slits that are each separated from the adjacent by a distance, d. When does constructive interference occur?

B.Find the angular positions of various orders of red (700 nm) and violet (400 nm) light passing through a grating with 6000 lines/cm (6000 slits in 1 cm).

Order / Violet Light, V / Red Light, R
m = 0
m = 1
m = 2
m = 3
m = 4
m = 5

C.View looking down at white light passing through a diffraction grating:

VI.Resolving Power for a Single Slit and a Circular Aperture

A.Diffraction pattern due to a slit and a circular aperture:

1.single slit

2.circular aperture

B.Look at the diffraction pattern produced by two noncoherent sources when the sources are “resolved” and “unresolved.”

Resolved:

Unresolved:

C.At some position, the sources go from being resolved to being unresolved. The criterion used is called Rayleigh’s Criterion: the center of the central maximum in one diffraction pattern coincides with the first minimum of the other diffraction pattern.

For slits:

For a circular aperture:

C.Examples

1.Two small sources of light whose dominant wavelength is , pass through a thin slit whose width is a. The light sources are separated by a distance d and are located a distance D from the slit. Their diffraction patterns are just resolved by Rayleigh’s criterion.

a.Draw the above situation.

b.What is their angular separation in terms of d, D, and .

2.Suppose you’re driving to Las Vegas on Interstate 15 at 11 in the evening and you see a car approaching you with their headlights on. Consider the pupil of your eye to be a circular with a diameter of 1.50 mm. The headlights of the car are 1.80 meters apart and emit a dominant wavelength of light of 550 nm. How far away is the car when the headlights are resolved by Rayleigh’s criterion?

3.The moon is 250,000 miles from the earth. How far apart can two points be on the moon when observed through the Mt. Palomar telescope so that they are resolved by Rayleigh’s criterion? (This essentially represents the smallest size an object can be in order to be seen by someone using the Mt. Palomar telescope. This ignores atmospheric effects that reduce its resolving poeer.) Take the dominant wavelength observed to be 500 nm, and let the Mt. Palomar telescope mirrorhave a diameter of 200 inches (= 508 cm).

1