EP025 Chapter 2 Physical Optics

CHAPTER 2 : PHYSICAL OPTICS
( 9 HOURS )

Terminology :

1. Huygen’s Principle : Prinsip Huygen.
Muka gelombang.
2. Wave front : Beza lintasan.
Interferens.
3. Path difference : Membina.
Membinasa.
4. Interference : Jalur.
Dwi-celah Young.
5. Constructive : Filem (saput).
Cincin Newton.
6. Destructive : Pembelauan.
Parut pembelauan.
7. Fringe : Monokromatik.

8. Young’s Double-slits :

9. Film :

10. Newton’s rings :

11. Diffraction :

12. Diffraction grating :

13. Monochromatic :

Introduction : JOTTING
SPACE
1. What is the nature of light ? Does it travel as a stream of particles or actually in a
form of certain types of waves ?

2. Properties of wave of light :
a. A form of energy (known as photon).
b. Propagates (spread) to all direction (3D).
c. Has constant velocity, c ≈ 3.0x108 ms-1 (1080 million kmh-1).
d. An element of electromagnetic waves.

3. Phenomena related to light :
a. Reflection.
b. Refraction.
c. Interference.
d. Diffraction.

All the phenomena can be explained physically by using a principle called
‘Huygens Principle’ which is based on the so-called ‘wave front’.

| Chapter 2 | Physical Optics | page 1/28 |

Huygen’s Principle :

1. Huygen’s principle states that :
“Each point on a wave-front acts as a secondary point source emitting spherical
wavelets. In a short time interval t, the new positions of the wave-front will be the
envelope of all the secondary wavelets.”

© Physics Teaching Courseware © Physics Teaching Courseware

2. Figure below shows the construction of Huygen’s principle. JOTTING
SPACE
r = ct

r = ct

wave-front •• •
• •
•
•

source

wave-front

λ λ

spherical wave-front plane wave-front

Note : Wave-front can be defined as a line or surface, in the path of a wave
motion, on which the disturbances at every point have the same phase.

The dark lines represent wave-front in a short time interval, t and the arrows to
show the direction of propagation. The line of wave-front is always perpendicular
with the direction of waves.

The radius of each wavelet is the distance between successive wave-fronts
represents the wavelength of the wave.

| Chapter 2 | Physical Optics | page 2/28 |

‘Plan-view’ :

direction of
wave propagation

‘Side-view’ : 360o 270o 180o 90o 0o degree (o)
2π 3π π π 0 radian (rad)
JOTTING
22SPACE

direction of wave
propagation

Explanation on diffraction patterns by using Huygen’s principle

1. Huygens’ principle can be used to explain
the diffraction of wave.

2. Each of the point in figure shown, acts as
a secondary source of wavelets (red circular
arc)

3. The tangent to the wavelets from points 2, 3
and 4 is a plane wave-front.

4. But at the edges, points 1 and 5 are the last
points that produce wavelets.

5. Huygens’ principle suggest that in conforming
to the curved shape of the wavelets near the
edges, the new wave-front bends or diffracts
around the edges - applied to all kinds of
waves.

© Mohd. Hazri @ kmph

© Mohd. Hazri @ kmph

If the size of the slit is small (a << λ),
then diffraction will occur as shown in the next figure.

| Chapter 2 | Physical Optics | page 3/28 |

Interference :

1. Interference occurs when ≥ 2 coherent waves overlap at a point.

Note : Coherent waves are waves that have the following properties :
i. same frequency.
ii. fixed phase difference.

2. Condition for interference :
i. The sources must be coherent.
ii. The superposition principle must be applied.

3. There are two types of interference :
i. Constructive interference.
ii. Destructive interference.

4. Type of interference depends on a value called ‘path difference, ∆L’.

Note : Path difference, ∆L refers to the difference of distance between two waves
from the respective sources, S1 and S2.

•C •D •C •D C• •D •C D• C• pattern JOTTING
•••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••••• SPACE
C = Constructive
screen D = Destructive

interference.

interference.

• ••••• •
S1 S2

Interference pattern

| Chapter 2 | Physical Optics | page 4/28 |

Constructive Interference :

1. Constructive interference at a point occurs if and only if the path difference, ∆L
between the waves has to be any whole number of wavelength, i.e. :

∆L = 0, λ, 2λ, 3λ, …..

Or ∆L = mλ …..(2.1)

where m = 0, 1, 2, 3, …

∆L : … 3λ 2λ λ 0 λ 2λ 3λ 4λ …

screen •C •C C• C• C• pattern
•••••••
••••••••••••• • ••••••••••••• S1 • JOTTING
S2 • SPACE
• ••••• •
© Physics Teaching Courseware
S1 S2

Destructive Interference :

1. Destructive interference at a point occurs if and only if the path difference, ∆L
between the waves is an odd multiple of λ/2 of wavelength, i.e. :

∆L = λ , 3λ , 5λ ,...
22 2

Or ∆L = m'+ 1 λ …..(2.2)
 2

where m’ = 0, 1, 2, 3, …

∆L : … 5λ 3λ λ λ 3λ 5λ …
222222

screen •D •D •D D• pattern
••••••••••••••••••••••••••••••
•• S1 •
S2 •
S1 S2
© Physics Teaching Courseware

| Chapter 2 | Physical Optics | page 5/28 |

Interference Fringes by a Double-slits :

1. Young experiment uses the Huygen’s principle to explain the phenomenon of
interference of light.

S2

S1

© Physics Teaching Courseware

S1 and S2 serve as coherent sources. Superposition between light waves from S1
and S2 produces interference patterns in the form of symmetrical bright and dark
stripes with the same thickness known as ‘interference fringes’.

2. Equations related to interference using Young’s Double-slits :

m m’ JOTTING
SPACE
4
3
3
2
y 2 1 (2nd dark)
1 0 (1st dark)
dQ 0
R O 0
1

21

2

Based on triangle OPQ : © Physics Teaching Courseware
tanθ = y
D …..(i)

Based on triangle S1S2R :

sinθ = ∆L …..(ii)
d

For θ → 0o, tanθ ≈ sinθ, then combination between equation (i) and (ii) :

y = ∆L or y = (∆L)D
Dd d

a) Equation for mth bright fringe :

ym = mλD …..(2.3)
d

where m = 0, 1, 2, 3, …

| Chapter 2 | Physical Optics | page 6/28 |

Note : Bright Bright Path Difference
Order Fringe (∆L)
Order
(m) 0th Central 0
1st 1st λ
0 2nd 2nd 2λ
1 3rd 3rd 3λ
2 … …
3 …
…

b) Equation for m’th dark fringe :

m'+ 1 λD
ym’ =  2
d …..(2.4)

where m’ = 0, 1, 2, 3, …

Note : Dark Dark Path Difference JOTTING
Order Fringe (∆L) SPACE
Order
(m’) 0th 1st λ /2
1st 2nd 3λ /2
0 2nd 3rd 5λ /2
1 3rd 4th 7λ /2
2 … …
3 …
…

3. The separation between successive (consecutive) bright or dark fringes, ∆y is
given by :

∆y = ym+1 - ym

= (m + 1)λD − mλD
dd

= λD
d

Note : Based ∆y depends on :
i) the wavelength of light, λ
ii) the distance apart, d of the double slits,
iii) distance between slits and the screen, D

| Chapter 2 | Physical Optics | page 7/28 |

4. Explanation for the above factors: JOTTING
(i) if λ is short and hence ∆y decreases for fixed D and d. The interference SPACE
fringes are closer to each other and vice-versa.
(ii) if the distance apart d of the slits diminished, ∆y increased for fixed D
and λ and vice-versa.
(iii) if D increases ∆y also increases for fixed λ and vice-versa.
(iv) if one of the slit, S1 or S2 is covered up, the fringes disappear.
(v) if the source slit is moved nearer the double slits, ∆y is unaffected but
their intensity increases.
(vi) if the experiment is carried out in a different medium, for example water,
the fringe separation ∆y decreased or increased depending on the
wavelength, λ of the medium.
(vii) if white light is used the central bright fringe is white, and the fringes on
either side are colored. Blue is the color nearer to the central fringe and
red is farther away as shown in figure below.

© Physics Teaching Courseware

| Chapter 2 | Physical Optics | page 8/28 |

Example : JOTTING
In a Young’s double slits experiment, the distance between the 8th order of bright fringes on SPACE
the two sides of the centre is 12.5 cm. If the separation between the slits and screen is 1.5
m and the wavelength is 550 nm, calculate the slit separation. [Ans: d = 1.056x10-4m]

Example :
Monochromatic light of wavelength 550 nm is used in a double-slits experiment with the slit
separation being 0.05 cm. Calculate the angle between the 5th order bright fringe to the 3rd
dark fringe. [Ans: ∆θ = 0.157o]

Interference in Thin Film :
1. A very common application of thin film interference is to measure the thickness, t

of very thin objects.
Note : Since the thickness of the film is uniform, then the pattern produced is

either bright or dark only (not in fringes)
2. Consider a parallel beam of light incident at angle α on a thin film of refractive

index n.

Thin film of
refractive index n

© Physics Teaching Courseware

3. Ray 1 and 2 serve as coherent sources. Superposition between the two waves
produces whether constructive interference or destructive interference depends on
their path difference in film, ∆Lfilm.

The path difference depends on :
a) the thickness, t of the film.
b) refractive index, n of the film.

4. Graphically (based on the above figure), the path difference between ray 1and 2 :

∆Lfilm = 2 t β 
cos

For normal incident, β → 0o, then cosβ → 1.

Thus, ∆Lfilm = 2t …..(i)

| Chapter 2 | Physical Optics | page 9/28 |

Note : The change of wavelength (or path difference) in different media :

Based on Snell’s law and Huygen’s principle :

λair λ air = nfilm where nair = 1.00
λ film nair

air Then :

film λ air = nfilm
λ film
λfilm

Or :

x(λair ) = nfilm
x(λfilm )

where x = m or x = m + ½

Hence :

∆Lair = nfilm or ∆L film = ∆Lair JOTTING
∆L film nair SPACE

Note : The phase change in different media :

Example 1 :
Wave incidents from less dense medium to denser medium.

Transmitted pulse

Incident pulse

Reflected pulse

Example 2 :
Wave incidents from denser medium to less dense medium.

Incident pulse Reflected pulse Transmitted pulse

Example 3 :
Wave incidents from a medium to another medium of the same dense.

Incident pulse No Reflected Transmitted pulse
pulse

| Chapter 2 | Physical Optics | page 10/28 |

Interference in Thin Film of refractive index 1n2n1 :

© Physics Teaching Courseware Medium of
JOTTINGrefractive index n1
SPACE
Thin film of
refractive index n2

Medium of refractive
index n3 < n2

1. Reflection at point A or D (ρ2 > ρ1) produces phase difference, ∆φ of π radian
(equivalents with λ/2).

There is no phase difference, ∆φ for reflection at C since ρ3 < ρ2.

2. Therefore, the path difference for constructive interference :

m + 1 λ
∆Lfilm =  2
n …..(ii)

where m = 0, 1, 2, 3, …

and the path difference for destructive interference :

∆Lfilm = mλ …..(iii)
n

where m’ = 0, 1, 2, 3, …

3. Based on equation (i), (ii) and (iii), the equations related to interference in Thin Film
of refractive index 1n2n1 :

a. Equation for mth bright order :

2nt = m + 1 λ …..(2.5)
 2

where m = 0, 1, 2, 3, …

b. Equation for m’ th dark order : …..(2.6)
2nt = m’λ

where m’ = 0, 1, 2, 3, …

Note : m’ = 0 refers to the occurrence of destructive interference where the
thickness of the film is too thin, t → 0 m (a layer of film atoms).

Normally, we take m’ = 1 to determine the minimum thickness of thin film.

| Chapter 2 | Physical Optics | page 11/28 |

Interference in Thin Film of refractive index 1n2n3 :

Medium of
refractive index n1

Thin film of
refractive index n2

C Medium of

refractive index n3
© Physics Teaching Courseware
JOTTING
SPACE
1. Reflection at A, D and C (ρ3 > ρ2 > ρ1) produce phase difference, ∆φ of π radian
respectively (equivalents with λ/2 respectively).

2. Therefore, the path difference for constructive interference :

∆Lfilm = mλ where m = 0, 1, 2, 3, …
n

and the path difference for destructive interference :

m + 1 λ
∆Lfilm =  2
n where m’ = 0, 1, 2, 3, …

3. Equations related to interference in Thin Film of refractive index 1n2n3 :
a. Equation for mth bright order :

2nt = mλ …..(2.7)

where m = 0, 1, 2, 3, …

Note : m = 0 refers to the occurrence of constructive interference where the
thickness of the film is too thin, t → 0 m (a layer of film atoms).

Normally, we take m’ = 1 to determine the minimum thickness of thin film.

b. Equation for m’ th dark order : …..(2.8)
2nt = m'+ 1 λ
 2

where m’ = 0,1, 2, 3, …

| Chapter 2 | Physical Optics | page 12/28 |

Example :
Light from a sodium lamp with wavelength of 589 nm is shone normally on a soap film of
refractive index 1.40. If the soap film appeared dark when viewed from the incident side,
determine the minimum thickness of the soap film. [Ans: tmin = 21.04 µm]

Interference in Thin Air Wedge :

© Physics Teaching Courseware
JOTTING
SPACE

m= 0 1 2 3 4 5 6 7 …
…
m' = 0 1 2 3 4 5 6 7 8

1st bright fringe

1st dark fringe

1. ABE and ABCDF serve as coherent sources. Superposition between the light

waves produces interference patterns in the form of bright and dark stripes with the

same thickness known as ‘interference fringes’ and depends on their path
difference, ∆L.

The path difference depends on the thickness, t of the air wedge.

2. Ray ABE does not change phase because it travels from a medium with a larger
refractive index to one with a smaller refractive index.

Ray ABCDF changes phase at C because it travels from a medium with smaller
refractive index to one with a bigger refractive index.

3. Therefore, the path difference for constructive interference :

∆L = m + 1 λ where m = 0, 1, 2, 3, …
 2

and the path difference for destructive interference :

∆L = m’λ where m’ = 0, 1, 2, 3, …

4. The path difference, ∆L between rays ABE and ABCDF is given by :

∆L = 2nt with nair = 1

Therefore, ∆L = 2t

| Chapter 2 | Physical Optics | page 13/28 |

5. Equations related to interference in Thin Air Wedge :
a. Equation for mth bright order :

2t = m + 1 λ …..(2.9)
 2 …..(2.10)

where m = 0, 1, 2, 3, …

Note : Bright Bright Path Difference
Order Fringe (∆L)
Order
(m) 0th 1st λ /2
1st 2nd 3λ /2
0 2nd 3rd 5λ /2
1 3rd 4th 7λ /2
2 … …
3 …
…

b. Equation for m’ th dark order :

2t = m’λ JOTTING
SPACE
where m’ = 0, 1, 2, 3, …

Note : Dark Dark Path Difference
Order Fringe (∆L)
Order
(m’) 0th 1st 0
1st 2nd λ
0 2nd 3rd 2λ
1 3rd 4th 3λ
2 … …
3 …
…

Example :

An air wedge is illuminated normally with monochromatic light of wavelength 560 nm.
Calculate the thickness of the wedge at the (a) 5th order bright fringe (b) 8th dark fringe.
[Ans: (a) t = 1.54 µm (b) t = 1.96 µm]

| Chapter 2 | Physical Optics | page 14/28 |

Newton’s Rings :

1. An interesting example of interference in thin film is the phenomenon of Newton’s
rings.
When a spherical surface lens is put in contact with an optically flat plate, the

circular fringes, known as ‘Newton’s rings’ are observed.

E

D © Mohd. Hazri @ kmph

© Mohd. Hazri @ kmph

2. Detail explanation for the formation of Newton’s rings is just the same as the JOTTING
formation of interference in air wedge. SPACE

3. If OD <<, then the path difference, ∆L between rays OL and OBDE is given by :

∆L = 2nt with nair = 1

Therefore, ∆L = 2t

4. Equations related to Newton’s rings :
a. Equation for mth bright order :

2t = m + 1 λ …..(2.11)
 2

where m = 0, 1, 2, 3, …

Note : Bright Bright Path Difference
Order Ring (∆L)
Order
(m) 0th 1st λ /2
1st 2nd 3λ /2
0 2nd 3rd 5λ /2
1 3rd 4th 7λ /2
2 … …
3 …
…

b. Equation for m’’th dark order :

2t = m’λ …..(2.12)

where m’ = 0, 1, 2, 3, …

| Chapter 2 | Physical Optics | page 15/28 |

Note : Dark Dark Path Difference
Order Ring (∆L)
Order
(m’) 0th Centre 0
1st 1st λ
0 2nd 2nd 2λ
1 3rd 3rd 3λ
2 … …
3 …
…

Note : The centre of the ring (contact point), O is dark (minimum)
because ray 1 is also undergoes a 180o (π rad) phase change with

respect to ray 2.

5. Diameter of the dark rings (approximate tm <<) :

dm’2 = 4 m' λ R
n

where m’ = 0, 1, 2, 3, …

n = 1.00 (in air)

JOTTING
SPACE

R R
R - tm R- tm

dm/2 dm/2
tm

Example :
In a Newton’s rings set up in air, the 13th dark ring has a diameter of 0.65 cm when
monochromatic light of wavelength 560 nm is used. Calculate the radius of curvature of the
curved surface of the plano-convex lens used. [Ans: R = 1.45m]

| Chapter 2 | Physical Optics | page 16/28 |

Diffraction :

1. Diffraction is a phenomenon refers to the spreading or bending of light waves over
the geometrical region as they pass through an obstacle or an aperture whose the
size of which are comparable to its wavelength.

λ

a Geometrical region
a~λ

© Physics Teaching Courseware

Note :
Diffraction produced by an obstacle.

JOTTING
SPACE

© Mohd. Hazri @ kmph

2. The widths of the central bright fringe (or also known as maximum) and the fringes
depend on the width of the slit : the smaller the slit width, the larger the widths of
the fringes.

© Physics Teaching Courseware

| Chapter 2 | Physical Optics | page 17/28 |

Diffraction by a Single-slit :

1. The thicknesses of dark fringe (or also known as minimum) are much shaper and
narrower than that of bright fringes. Therefore, the positions of dark fringes can be
determined more accurately.

2. Consider a plane monochromatic wave of a wavelength λ incident on a slit of a
width a.

θ1

aA P

a 2E yn
a C ∆L O (central maximum)

2B D
screen

λ JOTTING
SPACE
where ∆L = a sin θ1
2

- The slit is split into two equal parts, AC and CB. A, C and B are new
sources of secondary wavelets (based on the Huygen’s principle).

- When the wave-fronts from A, C and B superpose, interference will occur
at P. As AB is very small, thus AE is perpendicular to CP and AP = EP,
and therefore the path difference at P between ray AP and CP is :

Path difference, ∆L = CE

= AC sinθ1 …(based on trigonometric method)
…..(i)
= a sin θ1
2

- For the first minimum (1st order) at P :

Path difference, ∆L = λ …..(ii)
2

- From equation (i) and (ii) :

a sin θ1 = λ Or a sinθ1= λ
2 2

| Chapter 2 | Physical Optics | page 18/28 |

- For the second minimum (2nd order) at P : P

θ2 yn
O (central maximum)
aA D
a 4 C∆L E screen

B

2λ

Path difference, ∆L = λ …..(i)
2

- Based on the above figure, when the wave-fronts from A, C and B JOTTING
superpose, interference will occur at P. The path difference at P SPACE
between ray AP and CP is :

Path difference, ∆L = CE

= AC sinθ1 …(based on trigonometric method)
…..(ii)
= a sin θ 2
4

- From equation (i) and (ii) :

a sin θ 2 = λ Or a sinθ2= 2λ
4 2

- As a conclusion, all rays can be paired off and all the pairs result in
destructive interference. In general, the conditions for minimum can be
summarized as :

a sinθm’ = m’λ …..(2.13)

where m’ = 1, 2, 3, …

- If the distance of single slit to the screen is D, and D>>a then:

sin θm’ = tanθm’ = y m'
D

- The distance of nth order minimum from central maximum (zero-order
maximum):

ym’ = m' λD …..(2.14)
a

where m’ = 1,2, 3,…

| Chapter 2 | Physical Optics | page 19/28 |

Note : θ1 y1
θ1 y1
For first minimum (m’ = 1), the angle of diffraction
is θ1, but the angle 2θ1 is the angle subtended by 2θ1
the central maximum.

The width of central maximum is 2y1

To calculate the maximum number of orders
observed, then θ → 90o.

Note © Mohd. Hazri @ kmph

If we have two coherent sources, their images are not two point but two
diffraction patterns.

When the sources are close together, their diffraction patterns overlap. If they
are close enough, their patterns overlap almost completely and cannot de
distinguished.

S◘1◘S2 S◘1 S2 ◘S1 S2

◘ ◘

© Mohd. Hazri @ kmph
| Chapter 2 | Physical Optics | page 20/28 |

The effect of wavelength on the resolution of single slit :

A. One source :
Consider two light sources, SA and SA of wavelength λA (violet) and λB (red)
respectively where λB > λA, then the resolution of the diffraction pattern can be
described as follow :

θ1 y1 θ2 y1
y1 y1
θ1 θ1

S1 θ1 S2 θ1

2θ1

2θ1

Based on the equation of diffraction, it shows that : JOTTING
sinθ ∝ λ SPACE

Since λB > λA, then θB > θA.
Hence, the resolution of diffraction pattern produced by SB (red) is lower than
that of SA (violet).

Red λ increased → Resolution decreased.
Orange
Yellow

Green
Blue
Indigo
Violet

| Chapter 2 | Physical Optics | page 21/28 |

B. Two coherent sources :
Consider a pair of two coherent sources, SA1 and SA2 of wavelength λA (violet) and
SB1 and SB2 of wavelength λB (red) respectively where λB > λA and the separation
between the sources is the same. The resolution of the diffraction pattern can be
described as follow :

SA1

d

SA2

SB1 JOTTING
SPACE
d

SB2

It shows that the resolution of diffraction patterns produced by SB (red) is lower
than that of SA (violet).

Differences Between Diffraction and Interference Patterns :

Interference Diffraction
Pattern Pattern

The intensity for all bright fringes is The intensity for all bright fringes is
constant. varies. The closer the bright fringe
(maximum) to the centre, the brighter
All fringes (bright and dark) have the
same width. the fringe is.

All fringes have different width. The
closer the fringe to the centre, the
wider the fringe is. The central fringe

has the greatest wide.

| Chapter 2 | Physical Optics | page 22/28 |

Relationship Between
Resolution Power and Wavelength

(Zahidi, 2008)

Resolution
of Image

refers to…

The Sharpness
of an Image

depends on… reason… The Sharpness of an Image is Low
if the Diffraction Pattern is Obvious
Diffraction
Pattern (or Vise Versa)

depends on…

Glancing
Angle

depends on… JOTTING
SPACE
Magnitude
of Wavelength

Example :

A single slit diffraction pattern is formed on the screen which is placed 1.35 m from the slit.
The width of the slit is 0.11 mm and the 3rd dark band is formed at a distance of 14.5 mm
from the central maximum. Calculate the wavelength of the light used. [Ans: λ = 394nm]

Example :

A single slit diffraction pattern is obtained on a screen which is placed at a distance of 30
cm from the slit of width 7.3 µm. The wavelength of the light used is 650 nm. Calculate the
central bright fringe width (a) in degree (b) in cm. [Ans: (a) ∆θ = 10.2o (b) ∆y = 5.34cm]

| Chapter 2 | Physical Optics | page 23/28 |

Diffraction Grating :

1. In a single slit, the illumination on the screen is not uniform and consists of one
very bright ‘central maximum’ and several ‘secondary maxima’ which have
smaller intensities.

2. However, if we use more equally spaced slits, the bright fringes will remain at
the same places but they become sharper and narrower.

© Mohd. Hazri @ kmph

© Physics Teaching Courseware

© Physics Teaching CoursewareIf we wish to obtain very sharp maxima, which is necessary, for example in the
JOTTINGprecise determination of the wavelength of light, it is helpful to use a device
SPACEconsisting of a great number of narrow, closely spaced, equidistant and parallel
slits or lines. such a device is called a ‘diffraction grating’.

3. A diffraction grating containing slits is called a ‘transmission grating’. There is
also ‘reflection gratings’ in which fine lines are ruled on a metallic or glass
surface and light reflected from the lines is analysed.

Glass plate

d Line Slit

Slits between lines serve as coherent sources.

| Chapter 2 | Physical Optics | page 24/28 |

4. When there are N slits, then there are (N-1) ‘minima’ (dark fringes) between
each pair of principal maxima.

N’ = N-1
=8–1
=7

© Physics Teaching Courseware

3. The thicknesses of principal maxima are much shaper and narrower than that of
minima. Therefore, the positions of principal maxima can be determined accurately.

The condition for principal maxima is given by the following equation :

d sinθm = mλ …..(2.15) JOTTING
SPACE
where m = 0, 1, 2, 3, …

5. The number of grating lines per unit length, N is related to the distance between

slits, d and given by :

N= 1 …..(2.16)
d

6. Colourful bands from sunlight :
CD’s (compact dicks) are very common nowadays. They are used to store
information. But they can also act as reflection gratings due to the large amount
of closely spaced grooves on their surfaces.

If we illuminate a CD with sunlight or a lamp, beautiful bands of colour will appear
across its surface.

© Physics Teaching Courseware

7. Uses of diffraction grating :
a) To determine precisely the wavelength, λ of monochromatic light.
b) To describe the formation of white light spectrum.

| Chapter 2 | Physical Optics | page 25/28 |

8. Determining the wavelength, λ of monochromatic light using a diffraction grating :
When we illuminate a diffraction grating with a narrow beam of laser light,
several very bright principle maxima appear on the screen. Laser light is
monochromatic.

© Physics Teaching Courseware

The condition for the principle maxima is given by :
d sinθm = mλ

The expression for λ is thus given by : JOTTING
λ = d sin θ SPACE
m

where m = 0, 1, 2, 3, …
and θ can be calculate from the following condition :

θ = tan-1  y 
 D 

© Physics Teaching Courseware
| Chapter 2 | Physical Optics | page 26/28 |

JOTTING9. Describing the formation of white light spectrum :
SPACE
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White light actually consists of seven coloured spectrums which wove with same
speed, c but have different energy. Thus, they have different wavelength, λ and
frequency, f.
The arrangement of all the spectrum is as follow :

a) Purple, λP
b) Indigo, λI
c) Blue, λB
d) Green, λG λ increases
e) Yellow, λY
f) Orange, λO
g) Red, λR
Based on the condition for the principle maxima :

d sinθm = mλ
Since λR > λP, then θR > θP.
At the zeroth order (m = 0), all the spectrum diffracted at θ = 0o. Combinations of
all the spectrums form white light.

Example :
Calculate the number of line per mm on a diffraction grating if the 4th order maximum is
formed at an angle of 40.0o from the central maximum when monochromatic light of
wavelength 489 nm is used. [Ans: N = 328600m-1]

Example :
A laser light with a wavelength of 496 nm is projected normally upon a diffraction grating
with 4.9x105 lines per meter. (a) How many bright spots can be seen (b) how can you get
the number of bright spots to be more than in (a). [Ans: (a) 9 (b) …]

| Chapter 2 | Physical Optics | page 27/28 |

Table of Formula for Interference and Diffraction
(Zahidi, 2007)

Apparatus Formula Pattern
Young Double Slit
Bright Dark Symmetrical fringes.
Thin Film Central bright fringe, m =
(1n2n1) ym = mλD m'+ 1 λD
d ym’ =  2 0.
Thin Film d
(1n2n3) where m = 0, 1, 2, Either bright or dark.
where m’ = 0, 1, 2, For dark, tmin refers to m =
Air Wedge …
… 1.
Newton’s Rings
2nt = m + 1 λ 2nt = m’λ Either bright or dark.
Single Slit  2 where m’ = 0, 1, 2, For bright, tmin refers to m

Diffraction Grating where m = 0, 1, 2, … = 1.
…
Non-symmetrical fringes.
2nt = mλ 2nt = m'+ 1 λ 1st fringe refers to m = 0.
where m = 0, 1, 2,  2
Symmetrical rings.
… where m’ = 0, 1, 2, Central dark sport,
…
m =0.
2nt = m + 1 λ 2nt = m’λ JOTTING
 2 where m’ = 0, 1, 2, Symmetrical fringes. SPACE
The thickness of dark
where m = 0, 1, 2, … fringes is much narrower.
… 1st dark fringe refers to m

2nt = m + 1 λ 2nt = m’λ = 1.
 2 where m’ = 0, 1, 2, Symmetrical fringes.
The thickness of bright
where m = 0, 1, 2, … fringes is much narrower.
… Central bright fringe, m =
a sinθm’ = m’λ
- 0.
or

ym’ = m' λD
a

where m’ = 1, 2, …

d sinθm = mλ -
where m = 0, 1, 2,

…

| Chapter 2 | Physical Optics | page 28/28 |