Single-Slit Experiment
𝜽
𝒂
𝟒
𝜽
𝒂
𝟒
𝒂
sin 𝜽
𝟒
𝜽
𝒂
𝟒
𝒂
sin 𝜽
𝟒
𝜽
𝒂
𝟒
𝒂
sin 𝜽
𝟒
𝜽
𝒂
𝒂
sin 𝜽
𝟒
Single-Slit Experiment
• Therefore our general condition for destructive interference for a
single slit with width 𝑎 is:
𝒔𝒊𝒏 𝜽𝒅𝒂𝒓𝒌
a- Slit width
𝒏𝝀
=
𝒂
𝑛 = ±1, ±2, ±3, …
When the distance from slits to screen is much larger than the distance on the
screen, D >> y above, then the angle θ is "small" and we may assume tanθ is
approximately equal to sinθ which is approximately equal to θ (in radians).
For the position on the screen for maximum intensity.and we may write:
where:
 - wavelength ; d- slit serparation
D- distance to screen, y- distance separation
of the maxima
m or n - the order of the interference.
Double-slit interference
Double-slit interference
with light waves was
explored by Thomas
Young in 1801.

The following formula,
relates wavelength , slit
separation d, and distance
D to screen, to the
separation s of the maxima:
monochromatic
light
s = D / d
s
d
D
Young’s double-slit experiment
Diffraction Grating
- Recall and solve problems using the
formula d sinθ=nλ
- describe the use of a diffraction
grating to determine the wavelength of
light (the structure and use of the
spectrometer are not included);
Diffraction Grating
• It consists of a large number of equally spaced lines
ruled on a glass or plastic slide. Each line is capable
of diffracting the incident light.
• More slits mean
more interference
• Increases
brightness of
maxima
• Provides more
destructive
interference
DIFFRACTION GRATING
• For Constructive Interference (Maxima)
𝑑 sinϴ = nλ
• For Destructive Interference (Minima)
1
𝑑 sinϴ = (n + )λ
2
𝑑
• 𝑛 = 0, 1, 2, 3, 4… order
𝑑 sinϴ
= 𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒
where d is the distance between adjacent lines of the grating;
The distance d is known as the grating element or grating
𝑛=3
𝑑 sinϴ = nλ
𝑛=2
𝑛=1
𝜃
𝑛=0
𝑛=1
𝑛=2
𝑛=3
• it is usual to measure the angle at which they are
formed, rather than measuring their separation.
• With double slits, the fringes are equally spaced and
the angles are very small. With a diffraction grating,
the angles are much greater and the fringes are not
equally spaced.
Diffraction Grating
There may be as many as 10 000 lines per centimetre.
When light is shone through this grating, a pattern of
interference fringes is seen.
𝑠𝑙𝑖𝑡 𝑠𝑒𝑝𝑎𝑟𝑎𝑡𝑖𝑜𝑛
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
1 𝑚𝑚
100
𝟎. 𝟎𝟏 𝒎𝒎
1 𝑚𝑚
300
𝟎. 𝟎𝟎𝟑𝟑 𝒎𝒎
1 𝑚𝑚
600
𝟎. 𝟎𝟎𝟏𝟕 𝒎𝒎
EXAMPLE: A diffraction grating that has 750 lines per
millimeter is illuminated by a monochromatic light
which is normal to the grating. A third-order maximum
is observed at an angle of 56 to the straight-through
direction. Determine the wavelength of the light.
SOLUTION: Use n = d sin .
n = 3, and  = 56.
d must be calculated:
N = 750 lines / 1.0010-3 m = 750000 lines m-1.
d = 1 / N = 1 / 750000 = 1.3310-6 m.
  = d sin  / n = (1.33 10-6) sin 56 / 3
= 3.6810-7 m = 368 nm.
n=3
n=2
n=1
n=0
n=1
n=2
n=3
#Q2
Order of Interference
𝑛=3
𝑑 sinϴ = nλ
𝑛=2
𝑛=1
𝜃
𝑛=0
𝑛=1
𝑛=2
𝑛=3
• The greatest possible value of ϴ for diffraction is 90
degrees
• So to find the greatest order of diffraction, we take
the angle as 90 degrees. The value of sin 90 is 1.
Diffraction Grating
White Light
𝑛=2
𝑛 = 1 Red is
diffracted more
𝑛 = 0 than violet.
𝑛=1
𝑛=2
• A diffraction grating can be used to split
white light up into its constituent colours
(wavelengths).
• On either side, a series of spectra appear,
with violet closest to the centre and red
furthest away.
𝑛=2
𝑛=1
𝑛=0
𝑛=1
𝑛=2
• We can see why different wavelengths have their
maxima at different angles if we rearrange the equation
d sin θ = nλ to give:
sinθ = nλ/d
• The greater the wavelength λ, the greater the value of
sin θ and hence the greater the angle θ.
• Red light is at the long wavelength end of the visible
spectrum, and so it appears at the greatest angle.
Q10
Problem Solving
#Q3
Answer:
Answer