Diffraction II Review: Double Slit Path Differences

Diffraction II Review: Double Slit
Path Differences
• Today
For point P at angle θ triangle
– Single-slit diffraction review
– Multiple slit diffraction review shows ∆L = d sinθ
– Xray diffraction
– Diffraction intensities For constructive interference

we need ∆L = mλ

where m=0,1,2,… is any integer.

So the bright fringes are at

angles given by d sinθ = mλ

Double-slit interference fringes Single slit: Pattern on screen

So the bright fringes d sinθ = mλ Bright and dark fringes appear behind a single very thin slit.
are at angles given by As the slit is made narrower the pattern of fringes becomes wider.

And the dark fringes d sinθ = (m + 1 )λ
are at angles given by
2

Dark Fringes Single Slit dark
in Diffraction fringes

Destructive interference:

(a / 2m)sinθ = (λ / 2)

a sinθ = mλ

Don’t confuse this with the
condition for constructive
interference for two slits!

In fact, note that there is a dark fringe when the rays
from the top and bottom interfere constructively!

Summary of single-slit diffraction Example: Problem 36-6

• Given light of wavelength λ passing through a slit of The distance between the first and fifth minima of a single-
width a. slit diffraction pattern is 0.35 mm with the screen 40 cm
away from the slit, with light of wavelength 550 nm.
• There are dark fringes (diffraction minima) at
angles θ given by a sin θ = mλ where m is an integer. Find the slit width. sinθ ≈ tanθ = y / D

• Note this exactly the condition for constructive a sinθ = mλ sinθ = mλ / a
interference between the rays from the top and
bottom of the slit. sinθ5 − sinθ1 = 5λ − 1λ = 4λ
a a a
• Also note the pattern gets wider as the slit gets
narrower. y5 − y1 = D sinθ5 − D sinθ1 = 4λD
a
• The bright fringes are roughly half-way between the
dark fringes. (Not exactly but close enough.) a = 4λD = 4 × 550 × 10−9 × 0.4
y5 − y1 3.5 × 10−4

= 2.5 × 10−3 m = 2.5 mm

Q.36-1 Q.36-1 A slit of width 50 µm is used

A slit of width 50 µm is used with monochromatic light to form a
with monochromatic light to form a
diffraction pattern. The distance between diffraction pattern. The distance between
dark fringes on a distant screen is 4 mm. If
the slit width is increased to 100 µm, what dark fringes on a distant screen is 4 mm. If
will be the new distance between dark
fringes? the slit width is increased to 100 µm, what

Give your answer in mm. (In the range 0-9.) will be the new distance between dark

fringes?

Pattern size is inversely proportional to slit
size: 2 times slit width means (1/2) times the
distance between fringes. Answer: 2 mm.

Multiple-Slit Diffraction Double-Slit
Diffraction
Now we can finally put together our
interference and diffraction results to see a = slit width
what really happens with two or more slits. d = slit separation
θ = angle on screen
RESULT: We get the two-slit (or multiple-slit)
pattern as in chapter 35, but modified by the Bright fringes due to 2-slit
single-slit intensity as an envelope.
interference: θ = mλ / d
Instead of all peaks being of the same
height, they get weaker at larger angles. Zero due to diffraction:

θ = λ / a, 2λ / a, "

Double-slit diffraction Two-slit and one-slit patterns

2 slits of zero width Actual photograph:
1 slit of width a = 5λ (a) = two slits
(b)=one list covered
2 slits of width a = 5λ
(Figure 36-15 from text page 1003.)

Scaling of diffraction patterns Example:Sample Problem 36-5

Notice a common feature of interference Two slits: d=19.44 nm, a=4.05 µm, λ=405 nm.
and diffraction patterns: The large-scale
features of the pattern are determined by (a) How many bright
the small-scale regularities of the object, fringes within the
and vice-versa. central peak?

Holograms and X-ray diffraction patterns are examples. (b)How many in the
first side peak?

Example:Sample Problem 36-5 Q.36-2 When red laser light is

Two slits: d=19.44 nm, diffracted by two slits of equal width, there are
a=4.05 µm, λ=405 nm.
many closely spaced bands of light inside wider
Solution:
bands. What features of the slits determine the
One-slit:
separation x between the closely-spaced bands?
θ = mλ / a
= .10, .20, .30, .04, " x

Two-slit: (1) Width of the individual slits.
(2) Distance between the two slits.
θ = mλ / d = .02083m (3) Ratio of distance to width.
= .0208, .0416, .0625, .0833, .1042, .1250, " (4) None of the above: it’s more complicated.

Q.36-2 When red laser light is X-Rays

diffracted by two slits of equal width, there are •X-rays are just light waves with shorter
wavelengths and higher photon energies.
many closely spaced bands of light inside wider •Since X-ray wavelengths are comparable to
atomic sizes, they are perfect for studying atoms
bands. What features of the slits determine the and the arrangement of atoms in crystals.

separation x between the closely-spaced bands? X-ray crystallogrphy

x In this way, using xrays of known
wavelength we can measure the
(1) Width of the individual slits. distances between atoms in a
(2) Distance between the two slits. crystal and determine the crystal
(3) Ratio of distance to width. structure.
(4) None of the above: it’s more complicated.

X-Rays and Crystals

A crystal surface
acts like a
diffraction grating
for X rays.

Bragg condition
for a bright spot:

2d sinθ = mλ

Single-slit Intensity Phasors for Single Slit

•We know where to find the dark fringes in the Break up the slit into many tiny zones, giving many
single-slit pattern. But can we calculate the rays of light, which come together on the screen.
actual intensity at a general point?
∆φ = Phase difference between adjacent rays
•Yes, using the phasor method. Iθ =  sinα 2 φ = Phase difference between top and bottom rays
•Book gives result on page 998: Im  α Em = Amplitude at center = Sum of all phasors
 Eθ = Amplitude at angle θ, get from diagram


Here Iθ is the intensity at angle θ on the screen.

Im is the intensity at the central maximum.

The angle α= φ /2, and φ is the phase difference
between the rays from top and bottom of slit.

First Maximum and Minimum Intensity for Single Slit

Remember of course φ =  2π (a sinθ ) Em = Rφ Eθ = 2R sin(φ / 2)
the relation between λ
phase difference and φ =0 Eθ = 2 sin(φ / 2)
path difference Em φ

Eθ = Em Iθ = 4 sin2(φ / 2)
Im φ2

Eθ = 0 Which gives the textbook
φ = m(2π )
result: Iθ  sinα 2
Im  α
a sinθ = mλ = 
