Single slit diffraction Book page 364 - 367 ©cgrahamphysics.com 2016

• Assume paths of the two rays are parallel This is a good assumption if D >>> d

• 𝑃𝐷 = 𝑅2 − 𝑅1 = 𝑑𝑠𝑖𝑛 𝜃 since sin 𝜃 =𝑃𝐷

𝑑

• Constructive interference occurs, when PD = nλ

• nλ = 𝑑𝑠𝑖𝑛 𝜃 n = 0, 1, 2, 3, ….

• Destructive interference occurs when PD = 𝑛 +1

2λ

• 𝑛 +1

2λ = 𝑑𝑠𝑖𝑛 𝜃 n = 0, 1, 2, 3, …

©cgrahamphysics.com 2016

Review double slit Core

Rays are not parallel but converge

• We can still use the same mathematical approach

• sin 𝜃 =𝑃𝐷

𝑑

• tan 𝜃 =𝑦

𝐷

• If 𝜃 is small, sin 𝜃 ~ tan𝜃

•𝑃𝐷

𝑑=

𝑦

𝐷 PD =

𝑦𝑑

𝐷 where y = distance between adjacent nodal

lines, d = slit separation, D = distance between slit and screen ©cgrahamphysics.com 2016

For constructive interference

• PD = nλ

• 𝑛λ =𝑦𝑛𝑑

𝐷 or 𝑦𝑛 =

𝑛λ𝐷

𝑑

• Fringe separation:

𝑦 = 𝑋𝑛+1 − 𝑋𝑛 =λ𝐷

𝑑

𝑑 =λ𝐷

𝑦

©cgrahamphysics.com 2016

Requirements • Single slit diffraction occurs when a wave is incident on a

slit of about the same size as the wavelength

• The diffraction pattern consists of a series of bright and dark fringes

Note • Central maxima has twice the angular width

of the secondary maxima, each of which have same width

• Intensity falls significantly – intensity of 1st secondary maxima ~ 5% of 𝐼0

• 2 ~ 2% • 3 ~ 1% • No light reaches the center of a minima,

but going towards a maxima, the intensity gradually increases 𝐼 ∝ 𝐴2

• The narrower the slit width, the greater the amount of diffraction ©cgrahamphysics.com 2016

The single slit equation

• A slit maybe considered as a row of n point sources, close together and vibrating in phase, each source producing secondary circular wavelets

• At each point on the screen, resulting interference will be the sum of the contributions from each of the n point sources, taking into account the phase of each contribution

©cgrahamphysics.com 2016

Direction of incident waves

• When light passes through a narrow gap the range of point sources is restricted to the width of the gap

• Interference between wavelets causes cancellation in certain directions

• Consider wavelets arriving straight at the screen

• Arrive in phase

• Bright central maximum

• Constructive interference ©cgrahamphysics.com 2016

D

d

Consider waves arriving at angle 𝜃1 with PD λ

• The waves arrive in phase at the slit, but the wavelets leaving the slit at an angle will no longer be in phase

• This means the top ray travels PD further than bottom ray of λ, the

second ray travels PD of λ

2 and so on

• The difference in path between the top and middle ray is λ

2

destructive interference occurs

• In this way all wavelets in the top half of the slit will interfere destructively with wavelets from the lower part of the slit

• When the wavelets reach the screen, they will interfere destructively ©cgrahamphysics.com 2016

The first nodal line

• For n = 1 sin 𝜃1 =λ

2𝑑

2

=λ

𝑑

λ = d sin𝜃1 equation for minimum at 𝑃1 where d = slit width and θ = angle of PD

𝜽𝟏

©cgrahamphysics.com 2016

The second nodal line

• Consider a wider angle 𝜃2 with PD 2λ

• sin 𝜃2 =2λ

𝑑

• 2λ = 𝑑 sin 𝜃2

• 𝑚λ = 𝑑 sin 𝜃𝑚 single slit equation for destructive interference ©cgrahamphysics.com 2016

Constructive interference • Consider waves arriving at an angle 𝜃 with PD

3λ

2

• Pair rays from bottom third will cancel with rays from middle third, each pair canceling out because PD between

them is λ

2 ©cgrahamphysics.com 2016

Constructive interference PD = λ • Rays passing through top third have no matching pairs left

to cancel with

• These reach the screen as maximum: 3λ

2−

λ

2= λ

• sin 𝜃1 =3λ

2

𝑑 →

3λ

2= 𝑑 sin 𝜃1 for 1st order maximum, excluding

central maximum

©cgrahamphysics.com 2016

Higher order maximum

• Increasing the angle 𝜃 we can create a PD of 5λ

2,

7λ

2,….. up to 𝑚 +

1

2λ, m∈ 𝐼

• When the PD is an odd number multiple of λ

2 , higher

order maxima are produced

• 𝑚 +1

2λ = 𝑑𝑠𝑖𝑛 𝜃𝑚 constructive interference

©cgrahamphysics.com 2016

Remember

• The single slit and the double slit have opposite maxima and minima

• Double slit: 𝑚λ = 𝑑 sin 𝜃𝑚

This is maximum / constructive interference

• Single slit: 𝑚λ = 𝑑 sin 𝜃𝑚

This is minimum / destructive interference

©cgrahamphysics.com 2016

Example • A slit of width 2.0× 10−5𝑚 is illuminated by red light of

wavelength 620nm. At what angle does the third order minimum occur?

Solution

• 𝑚λ = 𝑑 sin 𝜃𝑚

• sin 𝜃 =𝑚λ

𝑑=

3×620×10−9

2.0×10−5 = 0.093

• 𝜃 = sin−1 0.093 = 5.330

©cgrahamphysics.com 2016

Basic observations • 𝑑(𝑠𝑙𝑖𝑡 𝑜𝑝𝑒𝑛𝑖𝑛𝑔)~λ

Geometrical shadow Diffraction pattern

a) Straight edge

b) Single long slit 𝑑~3λ

c) Circular aperture

d) Single long slit

𝑑~5λ

©cgrahamphysics.com 2016

Intensity of maxima

• Areas of min intensity at sin 𝜃 =λ

𝑑,

2λ

𝑑,

3λ

𝑑……..

• Areas of max intensity at sin 𝜃 = 0,3λ

2𝑑,

5λ

2𝑑,

7λ

2𝑑……..

Determined by slit width d

©cgrahamphysics.com 2016

Intensity • Intensity of light decreases for each successive bright area

since more of the “n” light sources interfere destructively

• To find the distance of the center point of the diffraction pattern

• tan 𝜃1 =𝑦1

𝐷

D d

©cgrahamphysics.com 2016

Finding an equation for interference

• sin 𝜃1 =𝑃𝐷

𝑑=

λ

𝑑

• For small angle sin 𝜃~ tan 𝜃 =𝑦1

𝐷

• For D >>> y

•𝑦1

𝐷=

λ

𝑑 λ=

𝑑𝑦1

𝐷

where d = slit width, y = distance from central point, D = distance from slit

d

PD=λ

©cgrahamphysics.com 2016

General equation

• Dark lines - sin 𝜃 =𝑛λ

𝑑

• Bright lines - sin 𝜃 =𝑛+

1

2λ

𝑑 for n = 1, 2, 3…..

• For 𝜃 = 0 there is a bright central maximum that is twice as wide as the others

• The equation λ=𝑑𝑦1

𝐷 can be used to

find λ predict dimensions of pattern predict positions of nodal lines

©cgrahamphysics.com 2016

Example • Light with wavelength 670nm passes through a single slit

with width 12𝜇𝑚. Viewed on screen, 30 cm away, find a) how wide is the central maximum in i)degree and ii)cm b) what is the separation of adjacent minima (excluding the pair on either side of the central maximum)

• Solution

• A)i) sin 𝜃 =𝑛λ

𝑑=

670×10−9

12×10−6 = 0.0558

𝜃 = sin−1 0.0558 = 3. 20 width = 2 x 3.2 = 6.40

ii) sin 𝜃 =𝑦

𝐷

y = D sin 𝜃= 0.30 x sin 3.2 = 1.67 × 10−2𝑚 ~ 3.4cm

• B) λ=𝑑𝑦1

𝐷 y=

𝐷λ

𝑑=

670×10−9×0.30

12×10−6 = 1.7𝑐𝑚 ©cgrahamphysics.com 2016

Note: effect of variations of d and λ • Central maxima is exactly twice the width of the separation

of adjacent nodal lines

• Diffraction pattern becomes more noticeable the narrower the slit becomes

• Narrower slit – lots of maxima of nearly same intensity

• Maxima very closely together

• Width of pattern ↑ in proportion to λ

• Wavelength ↓ inversely with width of slit

• If slit width is >>> λ, then central maxima is very small

©cgrahamphysics.com 2016

Note

• 𝜃 =𝑛λ

𝑑 where n = 1 is the position of first

minima

• Minima are not actually separated by equal distances

• However, for values 𝜃 < 100 it is a good approximation to consider minima to be equally spaced

©cgrahamphysics.com 2016

Summary

• Narrower slit effect more noticeable lots of maxima same intensity closer together

• To be noticeable λ~d 𝑦 ↑ 𝑎𝑠 λ ↑ maxima increases as 𝑑 ↓

𝑦 ↓ 𝑎𝑠 1

𝑑 𝑎𝑠 𝑑 ↑

• If d >>> λ than y is very small as D ↑ 𝑦 ↑

©cgrahamphysics.com 2016

Circular slits

• If the slit is not rectangular but circular, we can find the diffraction pattern by using the formula

𝜃 =1.22λ

𝑑

where d = diameter of aperture 𝜃 = in radians

• The angle 𝜃 locates the first dark fringe relative to the central bright region

©cgrahamphysics.com 2016

Single slit diffraction with white light

• A single slit illuminated by white light each component color has a specific λ max and min for each λ will be located at a different angle

• For a given slit width colors with long λ (red, orange, etc) will diffract more than color with short λ (blue, violet, etc)

• Resulting diffraction pattern will show all colors of rainbow with blue and violet nearer to central position

• Red will appear at greater angles

©cgrahamphysics.com 2016