Chapter 38 Diffraction Patterns and Polarization 38.2 Single-Slit Diffraction

• Wavelength of light >> the width of a slit.

• Each portion of the slit acts as a source of waves (Huygen’s principle).

• Light from one portion of the slit can interfere with light from another portion.

• The resultant intensity on the screen depends on the direction !.

• Diffraction can be described only with a wave model for

light. • Intensity of the single-slit diffraction pattern (see textbook):

I! = I0sin("asin! / # )"asin! / #

$

% &

'

( ) 2

Where I0 is the intensity at ! = 0 (the central maximum), a is the width of the slit, and " is the wavelength of light.

• The intensity minima (destructive interference) occur when

!a sin! /" = m! (m = ±1, ±2, ±3,…) Therefore, the condition for destructive interference is

sin! = m "a

(m = ±1, ±2, ±3,…)

• In most cases, use sin! # y/L to calculate y, the distance on the screen.

Example: Light of wavelength 580 nm is incident on a slit of width 0.3 mm. The observing screen is 2m from the slit (a) Find the positions of the first dark fringes.

" = 580 nm = 580 ! 10-9 m, a = 0.3 mm = 0.3 ! 10-3 m L = 2 m

The first dark fringes occurs at m = 1, sin! = "a

When ! is small, sin! # y/L

y = !La=(580 "10 #9m)(2m)

0.3"10 #3m= 3.87"10#3m

(b) Find the light intensity 2 mm from the center.

I = I0sin(!asin" / #)!asin" / #

$

% &

'

( ) 2

y = 2 mm = 2 ! 10-3 m sin! # y/L = (2 ! 10-3 m)/(2 m) = 1 ! 10-3

!a sin! /" = ! (0.3 ! 10-3 m)( 1 ! 10-3)/ 580 ! 10-9 m = 1.625

I2mm = I0sin(!asin" / # )!asin" / #

$

% &

'

( ) 2

= I0sin(1.625)1.625

$ % &

' ( ) 2

= 0.38I0

Intensity of double slit interference Patterns

An actual double slit interference pattern is resulted from (1) the diffraction due to the individual slits, and (2) the interference due to the waves coming from different slits.

The single-slit diffraction pattern acts as an “envelop” for a

double-slit interference pattern:

!

I" = I0 cos2(#d sin"

$) sin(#asin" /$)

#asin" /$% & '

( ) *

2

38.3 Resolution of Single-Slit and Circular Apertures

Rayleigh’s criterion: Two images formed by an aperture are just distinguishable if the central maximum of one image falls on the first minimum of another image. • For single slit diffraction, the first minimum occurs at the

angle sin! = "a

According to Rayleigh’s criterion, this is the smallest angular separation for which the two images are resolved. Thus, the limiting angle of resolution for a slit of width a is

sin!min "!min =#a

• The limiting angle of resolution for a circular aperture of

diameter D is !min = 1.22"D

Q: How to increase resolution (minimize !min)? Problem: At what distance could one distinguish two automobile headlight separated by 1.4 m? Assume a pupil diameter of 6 mm and yellow headlights (" = 580 nm). The index of refraction in eye is approximately 1.33. Solution:

s=1.4 m

L n=1.33

!min "n

The images of headlights are two diffraction disks on retina. According to Rayleigh criterion, the minimum

resolvable angle is !min = 1.22"nD

.

Since !n =!n

, " !min = 1.22"Dn

Also s/L # !min

L = s!min

=sDn1.22"

=(1.4m)(6#10$3m)(1.33)1.22(580 #10$9m)

= 1.58#104m= 15.8km

38.4 The Diffraction Grating • A diffraction grating

consists of a large number of equally spaced, identical slits, typically, several thousand lines per cm.

• The pattern is the result of

the combined effects of interference and diffraction.

• The condition for maxima in the

interference pattern is d sin! = m" (m = 0, ±1, ±2, ±3,…),

which is identical to two-slit pattern.

• The intensity maxima in a grating pattern are much sharper than that in a two-slit pattern.

• Like a prism, a diffraction grating can be used to separate

color spectrum from a white light. Example: A helium-neon laser ($ = 632.8 nm) is incident normally on a diffraction grating containing 6000 lines/cm. Find the angle of the second-order maximum. Solution: The space between slits d =(1/6000) cm = 1667 nm. For the second-order maximum (m = 2),

sin!2 =2"d=2(632.8nm)1667nm

= 0.76

!2 = 49.4°