Note Physics 2 SP025

Learning outcomes

At the end of this topic, students should be able to:

8.1 Huygen’s principle 8.2 Constructive and destructive 2
interferences
a) State Huygen’s principle (e.g.
spherical and plane wave fronts). a) Define coherence.

(Lecture : C1,CLO1,PLO1, MQF LOD1) b) State the conditions for
interference of light.
b) Sketch and explain the wave
front of light after passing c) State the conditions of
through a single slit and constructive and destructive
obstacle using Huygen’s
principle. interference.

(Lecture : C2,CLO1,PLO1, MQF LOD1) *Emphasise on the path difference and
(Tutorial : C4, PLO4, CTPS3, MQF LOD6) equivalence to phase difference

(Lecture : C1,CLO1,PLO1, MQF LOD1)

8.3 Interference of transmitted light 8.4 Interference of reflected light in
through double slits thin film

a) Use : a) Identify the occurrence of phase
change upon reflection.
i. Ym  mD for bright fringes (maxima)
*From lower to higher refractive index, phase
d change =  rad or path difference = ½λ

ii. (m  12)D for dark fringes (minima) (Lecture : C2,PLO1, MQF LOD1)
d
Ym  b) Describe with the aid of a
diagram the interference of light
where (m=0,±1, ±2, ±3…….) in the thin films at normal
incidence.
(Tutorial : C3, CLO3,PLO4, CTPS3, MQF LOD6)
(Lecture : C2,PLO1, MQF LOD1)
b) Use chy angdiDng and explain the 3
effect of any of the

variables.

(Tutorial : C3, CLO3,PLO4, CTPS3, MQF LOD6)

8.4 continued... i. constructive interference 2 nt  (m  1 ) 
2
c) Use the following equations ii. destructive interference 2 nt  m
for reflected light with no
phase difference (Tutorial : C3, CLO3,PLO4, CTPS3, MQF LOD6)
(nonreflective coating) :
e) Explain the application of thin
i. constructive interference 2 nt  m films (eg : solar panel, glass tint)

ii. destructive interference 2nt  (m 12) (Tutorial : C4, CLO3,PLO4, CTPS3, MQF LOD6)

(Tutorial : C3, CLO3,PLO4, CTPS3, MQF LOD6)

d) Use the following equation for
reflected light of phase

difference  rad (reflective

coating) :

4

8.5 Diffraction by a Single slit 8.6 Diffraction Grating

a) Define diffraction. a) Explain with the aid of
diagram the formation
(Lecture : C1,CLO1,PLO1, MQF LOD1) diffraction.

b) Explain with the aid of

diagram the diffraction of a (Lecture : C2,CLO1,PLO1, MQF LOD1)

single slit. b) Apply d sinn  n where

(Lecture : C2,CLO1,PLO1, MQF LOD1) 1
N
c) Use d 

i. Yn  nD for dark fringes (eg : transparent compact
a disc, muslin cloth,etc)
(mii.iYnn im(na a)12)D for bright fringes
(Tutorial : C4, CLO3,PLO4, CTPS3, MQF LOD6)

(maxima) where (n=0,±1, ±2,

±3…….) 5

(Tutorial : C3, CLO3,PLO4, CTPS3, MQF LOD6)

LO 8.1(a)

8.1 HUYGEN’S PRINCIPLE wavefront Line joining all point of
adjacent wave, e.g. A, B
Wavefront A and C or D,E and F forms
D wavefronts
a line or surface which joins all the adjacent
points which have the same phase in a B Ev
wave.
CF

Wave front always perpendicular to the circular
direction of wave propagation.
λ wave front
Types of wavefront:
• (1) plane 
• (2) Circular
ray

Plane λ
Wave
front

LO 8.1(a)

HUYGEN’S PRINCIPLE

Geometrical construction governing the propagation of wavefronts

A’ Q1

PA1 A’ Q1 Each point on an A s Q2
wavefront acts as a P1
P2 Q2 secondary point P2 Q3
source emitting
P3 Q3 spherical wavelets. source
After a time , the P3
P4 s B’Q4 new position of the
B wavefront will be the B P4
surface that it
tangential to all these B’ Q4
secondary wavelets.
Construction of
Construction of new wave new wave front for
front for a plane wave
ray a circular wave

LO 8.1(a) New New
wavefront wavefront
Wavelet source
point The wave front Old Wave front 8
at a later time t2
Old Wave is tangent to all
front the wavelets

Each of these
points is the
source of a
spherical
wavelet.



LO 8.1(b)

The limits of rays optics

LO 8.1(b)

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

1  Each of the point in Figure shown, acts as a
2 secondary source of wavelets (red circular arc)
3
4  The tangent to the wavelets from points 2, 3, 4
5 and 5 is a plane wave front.
6
Plane  But at the edges (points 1 and 6) the spherical
wave wavelets going around the edge of the slit into
front regions it would not otherwise reach.

 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. 10

LO 8.1(b)

Diffraction is most obvious when the gap is approximately the same size as the 11
wavelength of the wave

Huygen’s principle is consistent with diffraction

Diffraction is most obvious
when the size of the gap is
approximately the size of its
wavelength

Characteristic Comparison before Larger gap produces
and after diffraction small diffraction
Wavelength whilesmaller gap
Frequency Unchanged produces large
Speed diffraction
Direction of Unchanged
motion
Unchanged
Shape of the
wave Changes as the waves
spread out after
diffraction

Spread out more when
passing small obstacle
Spread less when
passing larger obstacle

LO 8.2

8.2 CONSTRUCTIVE INTERFERENCE AND DESTRUCTIVE
INTERFERENCE

o Interference is o when two light o the resultant Types of
defined as the waves meet at displacement Interference
effect of a point, a at any point is
interaction bright or a the vector
between two dark region sum of the
or more waves will be displacements
which produced in due to the two
overlaps or accordance to light waves.
superposed at the Principle
a point and at of Principle of
a particular superposition. Superposition
time from the
sources. Effect of
interference
Definition

LO 8.2 (a)

WHAT IS MEANT BY COHERENCE ?

COHERENCE means

the sources must have the same wavelength or frequency.
the sources must have a constant phase difference between them.

(Relative phases
are stable in time
and space)

LO 8.2(a)

TYPES OF COHERENT SOURCES

At starting point, S1 and S2
have no phase difference

Coherent In phase Both sources start with
sources Antiphase crest or trough

SS21 start with crest,
start with trough


At starting point, S1 and S2 λ
have a phase difference of 2
( equivalent to )

LO 8.2 (b)

CONDITIONS FOR INTERFERENCE OF LIGHT

Permanent (or stable) interference between two sources of light only

take place if

1. The sources must be coherent i.e., they must have the same frequency and

be always in phase with each other or have a constant phase difference.

2. The sources should be monochromatic—that is, of a single wavelength.

3. Light waves that are interfering must have the same or approximately of
amplitude to obtain total cancellation at minimum or to obtain a good

contrast at maximum.

4. The distance between the coherent sources should be as small as

possible of the light wavelength ( ).

LO 8.2(c)

PATH DIFFERENCE ( )

screen o Path difference ( ) is defined as
difference in distances traveled by two
light waves meeting at a point.

o It has a direct equivalence to phase

difference ( ) .

Path difference correspond to
phase difference of

L o The path difference decides whether
constructive or destructive interference
Path difference, = |S2b  S1b| occurs when they arrive at a point.
= |r2 –r1| = |

LO 8.2(c)

Conditions for constructive and destructive
interference

If S1, S2 were two coherent sources and start in phase :

+=

Conditions for constructive interference: +=

A bright light is observed at b if their path- Conditions for destructive interference:
difference, r2—r1 is differ by an integral A darkness is observed if the path difference
number of wavelengths: r2—r1 differ by a half-integral number of
Two light waves arrives at b are in phase and wavelengths :
give rise to constructive interference. Two light waves arrives at c are 180° out of
phase and give rise to destructive interference.

LO 8.2(c)

But not all sources are always in phase.
If S1, S2 were two coherent sources and start out of phase (equivalent to .

At starting point, S1 and
S2 have a phase
difference of in which

λ is equivalent to path

2 difference of

This initial difference between the two coherent sources need to be taken into consideration.

Thus the conditions for constructive interference will be:

The conditions for destructive interference will be:

LO 8.2(c)

SUMMARY CONDITIONS FOR CONSTRUCTIVE AND
DESTRUCTIVE INTERFERENCE

Two Coherent Conditions
sources
Constructive interference Destructive interference

{Bright or Maxima} {Dark or Minima}

In phase

Antiphase

LO 8.3(a)

8.3 INTERFERENCE OF TRANSMITTED LIGHT THROUGH
DOUBLE SLITS

❷ Cylindrical wave fronts spread out from slit ❻ Bright fringe is observed
So and reach S1 and S2 in phase because when constructive
both slits are equal distances from So.The interference occurs and
waves emerging from slits S1 and S2 are intensity is maximum.
therefore also always in phase. So S1 & S2
are coherent sources. ❺ center of the
pattern is a
❶ A monochromatic light So bright fringe.
source is placed behind
a narrow slit So, which
acts as point source.

❸ Waves spread out behind each slits. ❼ Dark fringe is observed
when destructive
❹ The waves interfere in the region where they interference occurs and
overlap, resulting in a pattern of bright and intensity is minimum.
dark bands on the screen.

LO 8.3 (a)

= 4 5th dark / 4th order min

= 4 4th Bright = 3 4th dark / 3rd order min

= 3 3th Bright = 2 3th dark / 2nd order min

= 2 2nd Bright / 2nd order max
S1 = 1 2nd dark / 1st order min

S2 = 1 1st Bright / 1st order max
= 0 1st dark / 0th order min

= 0 Central Bright / 0th order max

= −1 Distance of bright fringe
= −2
(maxima) from central bright :

= −3

wavelength of the light = −4 Distance of dark fringe
: separation between 2 slits
The bright fringes (minima) from central bright :
Distance from slits to screen are labeled by the
integer , starting at
Distance of from central the central bright
order number (

LO 8.3 (b)

Effect of Changing , or on (or
interference pattern)

y Refer :

y Click here to view
simulations
o Separation between two of double slit
consecutive bright interference.
fringes or dark fringes,
Variables Interference pattern

Increasing Fringes will
be farther
(move screen
farther from slits) apart

Shorter Fringes will
be closer
o also known as fringe width Decreasing together

or fringe Separation (slits move closer Fringes will
be farther
together)
apart

LO 8.3 (a) & (b) 1
=
Example 1 = 0,1, 2, 3, 4, 5,

A viewing screen is separated from a 2
double-slit source by 1.2 m. The distance
between the two slits is 0.20 mm. The third 3( )(1.2) center
bright fringe is 9.49 mm from the center line. 9.49 × 10 = 0.2 × 10
Calculate the
(a) wavelength of the light. = . × m
(b) distance of the third dark fringe from
= ( + ½) 1
the center.
(c) distance between adjacent bright = 0,1, 2, 3, 4, 5,

fringes. 2

1.2 m ( + ½)(5.27 × 10 )(1.2)
= 0.2 × 10

= . × m

distance between adjacent bright fringes refer to
fringe separation,

5.27 × 10 (1.2) = . × m
= = 0.2 × 10

LO 8.3 (a) & (b) Find that the fringe positions corresponding to these two

Example 2 wavelengths are

A light source emits 1
visible light of two =
wavelengths: 430 nm = 0,1, 2, 3, 4, 5,
and 510 nm. The source
is used in a double-slit 2
interference experiment
in which the distance center
from screen and
separation between slits For nm : ( )
are 1.50 m and 0.0250
mm respectively. Find the
separation distance ( ) m
between the third-order
bright fringes. For nm : ( )


( ) m

separation distance between the third-order bright fringes

= ( ) − ( ) = . × − . ×

= . × m = 1.40 cm

8.4 INTERFERENCE OF REFLECTED LIGHT IN THIN
FILM

A thin film is a thin layer of material that has a different index of refraction
than its surroundings.

Light reflected from the upper and lower surfaces of a thin film comes
together in the eye at P and undergoes interference.
Thin film interference is responsible for all of the colors that appear in soap bubbles
and oil slicks.

LO 8.4(a)

Occurrence of Phase change upon reflection

When light waves undergo reflection, phase incident wave
changes occur as follows: reflected wave transmitted wave
1. When light travels through a material with

a smaller refractive index toward
a material with a larger refractive
index (e.g., air to gasoline), reflected ray
undergoes rad phase change that

is equivalent to one-half of a wavelength.

Smaller refractive index Larger refractive index
(Less dense medium) (Denser medium)

LO 8.4(a)

2. When light travels from a larger incident wave n1
toward a smaller refractive index
(e.g., glass to air), there is no phase n2 transmitted wave
change upon reflection at the
reflected wave
boundary.

n2 n1

Larger refractive index Smaller refractive index
(Denser medium) (Less dense medium)

LO 8.4(a)

CHECK YOUR UNDERSTANDING

Is there any phase change upon reflection for ray 1 and 2?

A : Ray 1 : ______________
Ray 2 : ______________

B : Ray 1 : ______________
Ray 2 : ______________

C : Ray 1 : ______________
Ray 2 : ______________

LO 8.4(b)

o When monochromatic light (a single wavelength) strikes
the film nearly perpendicularly (Normal incidence).

o At the top surface of the film reflection occurs and
produces the light wave represented by ray 1.

o However, refraction also occurs, and some light enters
the film. Part of this light reflects from the bottom
surface of the film and passes back up through the film,
eventually reentering the air. Thus, a second light wave,
which is represented by ray 2, also exists.

o Moreover, this wave (ray 2), having traversed the film
twice, has traveled farther than ray 1.

o Because of the extra travel distance, there can be
interference between the two waves.

o If constructive interference occurs, an observer whose
eyes detect the superposition of waves 1 and 2 would
see a uniformly bright film.

o If destructive interference occurs, an observer would
see a uniformly dark film.

2 factors taking into consideration when dealing
with interference in thin film

optical path difference between 2 Occurrence of phase change upon
reflected rays. reflection

Ray 2 travel Light travels through a material with a
extra optical smaller refractive index (less dense)
distance of toward a material with a larger
2 compare refractive index (denser) undergoes a
to ray 1
 radian phase change (equivalents

to ½ ) upon reflection

Less dense Denser Less dense Denser

LO 8.4(c)

THIN FILM I : REFLECTED LIGHT WITH NO
PHASE DIFFERENCE

Light is partially      0 Point D very close to B (BC and CD
reflected at upper become straight line), thus Ray 2
and lower surfaces  rad phase  rad phase travels approximately further
of a thin film than ray 1.
(refractive index, ) change 1 2 change
Constructive interference:
Ray 1 has phase A
change. phase
Ray 2 has n1  1.0 Destructive interference:
change.
B D
Both reflected rays t : thickness of where
act as inphase n  1.5 thin film
coherent source and
meet at a point C Destructive interference occur for
produces interference
pattern n2  3.5 minimum nonzero thickness thus is

known as nonreflective

coating.

LO 8.4(d)

THIN FILM II : REFLECTED LIGHT OF PHASE
DIFFERENCE RAD

Light is partially     0   rad At normal incidence, point D very
reflected at upper close to B (BC & CD becomes
and lower surfaces  rad phase no phase straight line) thus Ray 2 travels
of a thin film approximately further than ray 1.
(refractive index, ) change 1 2 change
Constructive interference:
Ray 1 has radian A
phase change.
Ray 2 has no phase nair  1.0 D Destructive interference:
change.
B t : thickness of where
Both reflected rays thin film
(Ray 1 & 2) act as n  1.33
antiphase coherent Constructive interference occur for
source and meet at a C minimum nonzero thickness thus is
point produces
interference pattern. nair  1.0 known as reflective coating.

LO 8.4(c) & (d)

Thin film Interference : Summary

THIN FILM INTERFERENCE : SUMMARY

Keywords indicate constructive

interference : strongly reflected, appears
clearly seen, being observed, bright .

Keywords indicate destructive

interference : Non reflective, missing,
unseen, no light is found, dark

Equation Reflected light of Reflected light with
phase difference no phase difference

(Anti phase coherent) (In phase coherent)
Constructive
Destructive
Destructive
Constructive

Credit to : Mdm Anis, Mdm Azmira, Ms YapLZ, Ms Hayati, Sir Fauzi, Ms ChongYL (KMJs) & Sir LiewGH, (KMPh)

How to identify CROSS PAIR METHOD
the equation for
1. Draw a cross
a thin film 2. Label C D C D
easily ?
C – constructive interference ; D – destructive interference

3. Make categories (In phase ; Antiphase)
4. Place formulae

C In Phase D constructive

E.g : For the thin film shown below, red light is seen in the
reflection.

Anti C ∅ = − 0 =
Antiphase
D Phase

LO 8.4(c) & (d) phase change No phase change

Example 3 2 coherent sources

Light with a wavelength of antiphase Cross pair
598.3 nm is shone
normally on a soap film. If Air ( = 1.00) C In Phase D
the film has an index of
refraction = 1.40 and is Soap ( = 1.40) 2 = 2 = ( + ½)
suspended in the air ( =
1.00), find the minimum D Anti Phase C
thickness for which it
appears dark in reflected Air ( = 1.00)
light ?
Appears dark  destructive interference
Problem solving Stategy
① Visualize the thin film where
② Identify any phase change for both For minimum thickness, m = 1

reflected rays  IN or ANTI phase m
③ Refer keywords, identify is C or D ?
④ Use cross pair identify the equation
⑤ Solve

LO 8.4(c) & (d) phase change phase change

Example 4 2 coherent sources

Solar cells are often inphase Cross pair D
coated with a transparent C In Phase
thin film, such as silicon Air ( = 1.00)
monoxide (SiO ; = 1.45)
to minimize reflective SiO ( = 1.45) 2 = 2 = ( + ½)
losses. A silicon solar cell
( = 3.50) is coated with a D Anti Phase C
thin film of silicon
monoxide for this Solar cell ( = 3.50)
purpose. Assuming
normal incidence, Least reflection  destructive interference
determine the minimum （ ） where
thickness of the film that
will produce the least For minimum thickness, m = 0
reflection at a wavelength
of 552 nm. nm

LO 8.4(e) References : Physics for scientist and engineering with modern physics Giancoli, Prentice Hall pg. 880
https://en.wikipedia.org/wiki/Anti-reflective_coating, University Physics Pearson pg. 1199

IMPORTANT APPLICATION OF THIN FILMS:
ANTI REFLECTIVE COATING: SOLAR PANEL, GLASS TINT

1) A glass surface reflect about 4 percent of light incident upon it.

2) An antireflective or anti-reflection (AR) coating is a type of optical coating applied to the
surface of glass to reduce reflection and increase transmission of light. A single coating can
reduce total reflection from 4 percent to 1 percent of the incident light.

3) Examples include anti-glare coatings on corrective lenses and camera lens elements, and
antireflective coatings on solar panels

o The non-reflective coating (AR) allows more light
into the spectacles – the eye receives more light.

Another effect is that another observer can see the

wearer's eyes better, because less reflected light

from the spectacles is added to that reflected from
38
the wearer's eyes.

LO 8.4(e) References : 1) Physics for scientist and engineering with modern physics Giancoli, Prentice Hall pg. 880
2) https://en.wikipedia.org/wiki/Anti-reflective_coating, 3) University Physics Pearson pg. 1199

o In typical imaging systems, this improves the efficiency since
less light is lost due to reflection.

o In complex systems such as telescopes and microscopes the
reduction in reflections also improves the contrast of the
image by elimination of stray light. This is especially
important in planetary astronomy.

o AR coating (silicon monoxide, SiO, on silicon

photovoltaic solar cells ( = 3.5) help to reduce reflection and

increase amount of light reaches the solar cells and thus

increase its performance.

o Another advantage is that the glare from the glass will be

reduced. Some say this allows the panels to blend in more

easily with its surrounding.

o Also it allows the panels to be installed near airports (a 39

panel without anti reflective coating might blind a pilot..).

LO 8.4(e) References : 1) Physics for scientist and engineering with modern physics Giancoli, Prentice Hall pg. 880
2) https://en.wikipedia.org/wiki/Anti-reflective_coating, 3) University Physics Pearson pg. 1199

IMPORTANT APPLICATION OF THIN FILMS:
REFLECTIVE COATING

o Reflective coatings work the opposite way to antireflective coatings.
o E.g. a coating with reflective index 2.5 causes 38% of the incident of energy to be reflected

compared with 4% or so with no coating.

o By use of multi-layer coating, it is possible to achive nearly 100% of reflection for a particular
wavelengths.

o Some practical applications are Solar reflective coatings, heat reflective coating, infrared
“heat reflectors” in motion-picture projectors and astronauts' visors

A woman wears A thin layer of gold on Heat reflective coating when coated on
sunglasses featuring a an astronaut's helmet visor the roofing materials, reflects sunlight to
highly reflective optical fends off dangerous effects a greater extent and prevents the roofing
coating of solar radiation materials from getting heated up.
So the air below the roof never gets hot,
keeping the rooms cool and comfortable.
This prevents the buildings for a longer
period, extending the life of the buildings.
In air-conditioned areas, an attractive40
power saving of upto 40% can be noted
during peak summer.

LO 8.5(a)

8.5 DIFFRACTION BY A SINGLE SLIT

Diffraction o Assumptions
o Diffraction is the bending of 1. Slit size is small relative to

light waves as they travel the wavelength of light

around obstacles or pass (
through an narrow slit.
2. Screen is far away

3. Rays are parallel

Amount of bending depends on the
relative size of the wavelength of
light to the size of the slit. If the slit
is much larger than light’s
wavelength, the bending will be
almost unnoticeable. How ever of
the slit < the amount of
bending is considerable and easily
seen with naked eye.

LO 8.5(b) Location on screen o Fig.(a) consider monochromatic light of wavelength,
λ passing through a narrow slit of width, and

shines on a distant screen.

( ≫ ) Fig. (b) Greatly magnified view of slit
Fig.(a)

o According to Huygen’s principle, each point on The wavelets from each point on the initial
the wavefront at the slit acts as secondary wave front overlap and interfere, creating a
sources emitting new wavelets that radiate diffraction pattern on the screen
towards the screen.

o Wavelets from one portion of the slit interfere
with wavelets from another portion and
generates diffraction pattern on screen.

LO 8.5(b)

Each point on the
wave front is
paired with another
point distance



away

o For , the wavelets going straight o These wavelets all meet on the screen at angle
The path difference between rays 1 and 2, rays 3
forward all travel the same distance to the
and 4 or rays 5 and 6 is (a/2) sin . If this path
screen. Thus they arrive in phase and difference is exactly half a wavelength

interfere constructively produces a bright (corresponding to a phase difference of 180°), then
the two waves cancel each other and destructive
central fringe on the screen directly interference results. A dark fringe is seen on

opposite the slit. screen.

DIFFRACTION PATTERN FROM A SINGLE SLIT

LO 8.5(b)

A SINGLE SLIT DIFFRACTION PATTERN

The brightness of 3 2 ima 2 ima Between each pair of
the fringe is related 3 ima dark fringes there is
ima 1 ima a bright fringe due to
to light intensity, 1 ima constructive
interference.
Most of the light
intensity is The higher order
bright fringes are
concentrated much less intense
in the broad
central bright than the central fringe as
fringe. the graph indicates.

LO 8.5(c)

FORMULAE FOR DIFFRACTION BY A SINGLE SLIT

Light intensity Diffraction o A dark fringe (minima) is obtained
on screen pattern at angle on either side of the

2nd maximum midpoint on the screen where

1st dark fringe 2nd minimum

(minima) 1st maximum

● 1st minimum width of the single slit
: angle for the nth dark fringe
Central
● maximum (minima) measured from the

Midpoint of ● 1st minimum midpoint on the screen
central 1st maximum
: wavelength of the monochromatic
maximum 2nd minimum light used

2nd maximum

: order number

LO 8.5(c)

FORMULAE FOR DIFFRACTION BY A SINGLE SLIT

Distance of order dark
fringe (minima) to the midpoint
of the central bright fringe,

minima where



1 minima : distance between single slit

& screen
1 minima Distance of order bright
fringe (maxima) to the midpoint : width of the single slit
of the central bright fringe, wavelength
: order number

Here to view single slit Width of the central bright
interference simulations
width of the central bright
goes from the first
minimum on one side to
the first minimum on the
= 1 other side




= 1

The width of the central bright is
inversely proportional to the width
of the slit, . The narrower the slit,
the more the waves spread out.

LO 8.5(c)

width of the central bright is directly proportional to the

wavelength, 

455 nm



577 nm



780 nm

LO 8.5(c)

Example 5 (c) Calculate the angle for
the third minima to be
Light of wavelength 580 observed.
nm is incident on a slit of
width 0.003 mm. The m
observing screen is 2 m
from the slit. Find (a) the (b) Calculate the width of
position of the first dark the central bright fringe.
fringe.

Given : m
= 0.003 m
= 1.6 m

Find (a)

m

LO 8.5(c) Given : 1 m (b) What will happen to
=1 m the diffraction
Example 6 maximum width if the
= 2.4 m m screen is moved
When a 500 nm light If 2 closer to the slit?
illuminated onto a single Find Explain your answer.
slit, a diffraction Apply :
maximum of width 10 
cm is formed on a 1
screen 2.4 m away from diffraction maximum
the slit. m width increases.
diffraction maximum
(a) Calculate the width
of the diffraction becomes broader.
maximum if the 500
nm light is replaced
by a 650 nm light.

m