Interference of Light
Contents
1.
2.
3.
4.
5.
6.
7.
8.
The nature of Light.
Corpuscular theory of Light
Principle of Superposition Light
Definition of Interference of Light.
Conditions for Interference.
Constructive and destructive interference.
Coherent Source.
Young’s double slit Experiment and finding the condition
for constructive and destructive interference.
9. Equation for the position of Bright and dark fringes. Hence
the separation between two consecutive bright or dark
fringes.
10. Intensity Distribution of the double-slit interference pattern
11. Phasor addition of waves
12. Multi -slit interference pattern
13. Producing interference pattern using (i) Lloyd’s mirror and
(ii) Fresnel’s biprism.
14. Observation of Interference effects in (i) thin films and
(ii) Newton's rings
15. Michelson Interferometer and its use in modern
Technology.
The nature of Light
(i) Wave nature (electromagnetic wave) and
(ii) Particle nature (bundles of energy called photons)
The two most successful theories of light were the corpuscular
(or particle) theory of Sir Isaac Newton and the wave theory of
Christian Huygens.
• Corpuscular theory of Newton (1670)
• Light corpuscles have mass and travel at extremely high speeds
in straight lines
• Huygens (1680)
• Huygens argued that if light were made of particles, when light
beams crossed, the particles would collide and cancel each
other. He proposed that light was a wave.
Wavelets-each point on a wavefront acts as a source for the
next wavefront.
The nature of Light (Cont..)
• Huygen assumed that light is a form of wave motion rather
than a stream of particles.
• Huygenʼs Principle is a geometric construction for determining
the position of a new wave at some point based on the
knowledge of the wave front that preceded it.
• All points on a given wave front are taken as point sources for
the production of spherical secondary waves, called wavelets,
which propagate in the forward direction with speeds
characteristic of waves in that medium.
• After some time has elapsed, the new position of the wave
front is the surface tangent to the wavelets.
• As you might expect, the heuristic idea of Huygens can be fully
justified through various derivations associated with the
Maxwell equations.
The nature of Light (Cont..)
The nature of Light (Cont..)
The nature of Light (Cont..)
• At the end of the 19th century, James Clerk Maxwell
combined electricity, magnetism, and light into one theory. He
called his theory the electromagnetic theory of light.
• According to Maxwell, light was an electromagnetic wave with
the same properties as other electromagnetic waves.
Maxwell’s theory, however, was unable to explain the
photoelectric effect.
• In 1900, Max Planck suggested that light was transmitted and
absorbed in small bundles of energy called “quanta.”
• Albert Einstein agreed with Planck’s theory and explained the
photoelectric effect using a particle model of light. The
quantum theory combines the two major theories of light,
suggesting that light does not always behave as a particle and
light does not always behave as a wave.
The nature of Light (Cont..)
Proofs of Wave Nature
• Young's Double Slit Experiment (1807)
bright
(constructive)
and
dark
(destructive) fringes seen on screen.
• Thin Film Interference Patterns.
• Poisson/Arago Spot (1820).
The nature of Light (Cont..)
• Diffraction fringes seen within and around a small obstacle
or through a narrow opening.
The nature of Light (Cont..)
Proof of Particle Nature
The Photoelectric Effect (PEE)
• Albert Einstein 1905
• Light energy is quantized
• Photon is a quantum or packet of energy
• Heinrich Hertz first observed the PEE in 1887
• Einstein explained it in 1905 and won the Nobel
prize for this.
Compton Effect
The scattering of photons from charged
particles is called Compton scattering
after Arthur Compton who was the first
to measure photon-electron scattering in
1922.
Proof of Particle Nature
• The conversion of a photon
into an electron-positron pair
on its interaction with the
strong electric field
surrounding a nucleus is called
pair production.
• The converse of pair
production in which an
electron and positron combine
to produce two photons is
known as annihilation of
matter.
• It is one of the principal ways in
which high-energy gamma
rays are absorbed in matter.
Superposition principle
The superposition principle of
light waves states that when
two or more waves overlap in
space,
the
resultant
disturbance is equal to the
algebraic sum of the individual
disturbances.
Coherent sources
Two narrow sources of light are said to
be coherent if they emit waves having. the same
wavelenght (or frequency), the same amplitude,
and. a constant phase relation between them.
How to make Coherent Sources in Laboratory?
Interference of light
When two light waves from different coherent sources meet
together, then the distribution of energy due to one wave is
disturbed by the other. This modification in the distribution of
light energy due to super- position of two light waves is called
"Interference of light".
Conditions for interference
➢ Light must be monochromatic, i.e., involve just a single
frequency (single wavelength), i.e. the two sources of light
should emit continuous waves of same wavelength and same
time period.
➢ Light sources must be coherent, the relative phase is always
the same, i.e. the waves emitted by two sources should either
have zero phase difference or no phase difference.
➢ Light sources must have the same amplitudes.
If these conditions do not hold, one still gets constructive and
destructive interference but the interference pattern can
change with time or not be complete (destructive interference
leads to a decrease in amplitude but not to zero amplitude).
Types of interference
Constructive interference
When two light waves superpose with each other in such away
that two interfering waves have a displacement in the same
direction so that the crest of one wave falls on the crest of the
second wave, and trough of one wave falls on the trough of the
second wave, then the resultant wave has larger amplitude than
each which is equal to their sum and it is called constructive
interference.
Destructive interference
When two light waves interfering with each other in such away
that they have a displacement in the opposite direction so that
the resultant displacement is minimum, this is known as
destructive interference. Destructive interference has the
tendency to decrease the resulting amount of displacement of
the medium.
Young's interference experiment, also known as the double-slit interferometer, was
the original version of the modern double-slit experiment, performed at the
beginning of the nineteenth century by Thomas Young (1773-1829) a British scientist.
This experiment played a major role in the general acceptance of the wave theory of
light.
Why Interference Patterns Are Not Standing Waves?
The interference pattern in Figure
shows bright and dark regions that
appear similar to the antinodes
and nodes of a standing-wave
pattern on a string. While both
patterns depend on the principle
of superposition, here are two
major differences:
(1) waves on a string propagate in
only one dimension while the
light-wave
interference
pattern exists in three
dimensions;
(2) the standing-wave pattern
represents no net energy flow,
while there is a net energy
flow from the slits to the
screen in an interference
pattern.
Young’s Double-Slit Experiment
From Young’s Double-Slit Experiment setup one can have:
i)
Describing the Young’s Double-Slit Experiment setup.
ii) Derivation of the equation for the condition for
constructive and destructive interference
iii) Explanation of the assumption L>>d, and d>>, where the
terms have their usual meaning.
iv) Finding the equation for the position of the bright and dark
fringes.
v) Finding that (a) the difference between the consecutive
𝑳
bright or dark fringes are same and is given by 𝜷 =
and
𝒅
(b) the difference between the consecutive bright and dark
𝑳
fringes is 𝞒 =
𝟐𝒅
Young’s Double-Slit Experiment (Cont…)
• In May of 1801, Young performed the double-slit
experiment to demonstrate the interference of light waves.
• The demonstration would provide solid evidence that light
was a wave, not a particle.
• Measured the wavelength of light.
Young’s Double-Slit Experiment (Cont…)
• The slit-width (a) and slit-separation (d) are similar in size to
the wavelength of light ()
• The wave fronts arrive at the two slits from the same source in
about the same time - they are in phase ( = 0)
• Slits S1 and S2 acts like point-sources by Huygen’s principle.
Young’s Double-Slit Experiment (Cont…)
• Slits S1 and S2 acting as coherent source with nearly same
amplitude.
• Light waves from S1 and S2 travelling the distances r1 and r2
interfere at P on the Screen. The screen is placed at a distance
L from the Slit axis.
•  is the angle corresponding to the point P at the slits
midpoint Q.
Young’s Double-Slit Experiment (Cont…)
Assumption that L>>d and <<d,
Under these conditions  is small; thus, we can use the small
angle approximation sin  tan
Young’s Double-Slit Experiment (Cont…)
The fringe spacing or fringe width for the bright fringe is
𝛽 = 𝑦𝑚+1 − 𝑦𝑚 =
𝑚+1 𝜆𝐿
𝑑
𝑚𝜆𝐿
−
𝑑
=
𝜆𝐿
𝑑
Similarly, the fringe spacing or fringe width for the dark fringe
𝛽′ = 𝑦′𝑚+1 − 𝑦′𝑚 =
[2(𝑚+1)+1]𝜆𝐿
2𝑑
−
[2𝑚+1]𝜆𝐿
2𝑑
=
𝜆𝐿
𝑑
Hence the fringe spacing for consecutive bright or dark fringes
𝜆𝐿
are same and is given as 𝛽 =
𝑑
Centre spacing between two consecutive bright and dark fringes
is
𝞒 = 𝑦𝑚 − 𝑦 ′ 𝑚 =
𝑚𝜆𝐿
𝑑
−
2𝑚+1 𝜆𝐿
2𝑑
=
𝜆𝐿
2𝑑
If distance d between slits is decreased, then the angles corresponding to the bright
fringes will remain unchanged but the fringes will all become brighter.
If wavelength λ of monochromatic light impinging on two-slits experiment increases,
then bright fringes all spread further apart.
Young’s Double-Slit Fringe Intensity
1. In Derivation of Young’s Double-Slit Experiment, We discussed
the locations of only the centers of the bright and dark fringes
on a distant screen.
2. we now calculate the distribution of light intensity associated
with the double-slit interference pattern.
3. Let us suppose that the two slits represent coherent sources
of sinusoidal waves such that the two waves from the slits
have the same angular frequency  and a constant phase
difference .
4. Assuming that the two waves have the same amplitude E0, we
can write the magnitude of the electric field at point P due to
each wave separately as
E1 = E0sint and
E2 = E0sin(t+ )
(1)
5. Although the waves are in phase at the slits, their phase
difference  at P depends on the path difference  = r2-r1 = dsin.
Young’s Double-Slit Fringe Intensity (Cont…)
6. A path-length difference of one wavelength () produces a
phase difference of 2 rad, which is equivalent to no phase
difference at all.
7. A path difference of  is the same fraction of  as the phase
difference  is of 2. We can describe this mathematically

𝛿
=
(2)
 2
Which gives us
2
2
 = 𝛿 = 𝑑𝑠𝑖𝑛
(3)


Young’s Double-Slit Fringe Intensity (Cont…)
Young’s Double-Slit Fringe Intensity (Cont…)
Young’s Double-Slit Fringe Intensity (Cont…)
Phasor Addition of Waves
Let us again consider a sinusoidal wave whose electric field
component is given by E1 = E0sint and
E2 = E0sin(t+ )
Phasor Diagrams for Two Coherent Sources
Multiple-slit interference patterns
As N, the number of slits,
is increased, the primary
maxima
(the
tallest
peaks in each graph)
become narrower but
remain fixed in position
and the number of
secondary
maxima
increases. For any value
of N, the decrease in
intensity in maxima to
the left and right of the
central
maximum,
indicated by the blue
dashed arcs, is due to
diffraction patterns from
the individual slits.
Producing interference pattern
There are different methods for producing coherent sources; but
all these methods may be conveniently classified into two main
categories.
Division of Wavefront:
(i) Lloyd’s single mirror, (ii) Fresnel’s biprism, (iii) Fresnel’s
double mirror, (iv) Billet’s divided lens and (v) Rayleigh’s
interferometer
Division of Amplitude:
(i) Thin Films, (ii) Newton’s rings, (iii) Michelson’s interferometer,
(iv) Jamin’s interferometer and (v) Fabry-perot interferometer
Lloyd’s Single Mirror
Humphrey Lloyd A light beam after reflection from an optically
Developed in 1834 denser medium undergoes a phase change of
 thus the path difference /2
• From the light source, ray 1 travels directly while the ray 2 is
reflected from the mirror surface and interfere with 1 at P on a
vertical screen.
• The reflected ray can be treated as a ray originating from a
virtual source at point S’.
• As the rays are reflected from the whole surface of the mirror,
its seems that light comes from the coherent source S’ and
hence fringe formed in the zone MN.
Fringe width MN
Lloyd’s Single Mirror (Cont…)
• Separation between the slits = SS = d. The screen Distance
very large in comparison of the slit separation (D>>d) and the
wavelength is very small (<<d).
• An interference pattern is indeed observed. However, the
positions of the dark and bright fringes are reversed relative
to the pattern created by two real coherent sources (Young’s
experiment). This occurs because the coherent sources at
points S and S’ differ in phase by 180° which in turn results a

path extra path difference .
2
• Interference at a distance on the screen, OP = x. At P the path
𝑑.𝑥
𝜆
+
𝐷
2
𝑑.𝑥
𝜆
+ =
𝐷
2
difference between the rays 1 and 2 is ∆ =
• For the Nth bright Fringe at P we use ∆ =
From which 𝑥𝑁𝐵 =
(2𝑁−1)𝜆𝐷
2𝑑
𝑁𝜆
Lloyd’s Single Mirror (Cont…)
• For the Nth dark Fringe at P we use ∆ =
From which 𝑥𝑁𝐷 =
𝑑.𝑥
𝐷
𝜆
2
+ = (2𝑁 + 1)
𝜆
2
𝑁𝜆𝐷
𝑑
Fringe width for both bright and dark, 𝛽 =
𝜆𝐷
𝑑
Why central point on the screen is dark instead of being bright?
• The central point O on the screen is equidistant from points S
and S’. The direct beam from S cannot under go a phase change
but the reflected ray has a 180 phase change from the mirror
surface. The refractive index of the mirror (medium 2) higher
than the medium 1 (Air).
• This can be explained based on the string analogy. The reflected
pulse on a string undergoes a phase change of 180° when
reflected from the boundary of a denser medium, but no phase
change occurs when the pulse is reflected from the boundary of
a less dense medium.
Lloyd’s Single Mirror (Cont…)
• Similarly, an electromagnetic wave undergoes a 180° phase
change when reflected from a boundary leading to an optically
denser medium (defined as a medium with a higher index of
refraction), but no phase change occurs when the wave is
reflected from a boundary leading to a less dense medium.
Fresnel’s Biprism
• Augustin-Jean Fresnel was French Physicist
Contributed
significantly
to
the
establishment of the wave theory of light
and optics.
• He gave a simple arrangement for the
production of Interference Pattern.
• Biprism consists of two identical thin prism of very small
refracting angle (30’ to 1) with their bases joined together.
• Thus the biprism BP is a thin glass prism of obtuse angle 179
and acute angle  is about 30 on both sides.
• A monochromatic light of wavelength  from a source passes
through the narrow slit S.
• The biprism is placed in front of the slit with its refracting
edge parallel to slit.
Fresnel’s Biprism (Cont…)
• The light from S is allowed to fall symmetrically on the biprism
BP. The light beams emerging from the upper and lower
halves of the prism appears to come from two virtual sources
S1 and S2 which act as coherent sources.
• The cones of light BS1E and AS2C, diverging from S1 and S2 are
superposed and the interference fringes are formed in the
overlapping region BC of the Screen.
Fresnel’s Biprism (Cont…)
• Let us suppose the distance between S and S is d.
• Moreover, the two emergent wavefronts intersect at small
angles and hence the fundamental condition of interference,
i.e. sin  tan approximation is satisfied.
• Interference fringes of equal width are produced in the
overlapping region BC of the screen but beyond B and C,
fringes of large width are produced due to Diffraction.
• The midpoint O of BC is equidistant from S and S, hence this
point has maximum intensity.
• Fresnel’s Biprism working mechanism totally same as that of
the Youngs-double slit experiment. So, on both sides of O,
alternately bright and dark fringes are produced.
Fresnel’s Biprism (Cont…)
• For the Nth bright fringe, the distance,
𝑥𝐵𝑟𝑖𝑔ℎ𝑡 =
𝑁𝜆𝐷
,
𝑑
N = 0, 1, 2, 3, ….etc
• For the Nth dark fringe, the distance,
𝑥𝑑𝑎𝑟𝑘 =
(2𝑁+1)𝜆𝐷
,
2𝑑
N = 0, 1, 2, 3, ….etc
• Hence the fringe width for both bright and dark, 𝛽 =
𝜆𝐷
𝑑
Advantages Fresnel’s biprism
➢ For the Young’s slits experiment, we must approximate that
the slits act as point sources. This however is not the case,
since the slits have finite width. This finite size of the
secondary slits gives rise to unwanted diffraction effects which
causes errors. The Fresnel biprism overcomes this problem of
extended secondary slits by replacing them with virtual slits
which are point-like.
Fresnel’s Biprism (Cont…)
➢ As this experiment is analogous to Young slits, the formula
above holds with the minor exception that d can not be
measured directly since the two slits are purely virtual.
➢ Instead, d is determined by placing a converging lens between
the biprism and the screen and forming real images of the
virtual slits on the screen. From the optical magnification
formula.
Interference effects by in thin films
Interference effects in Bubbles
Interference effects by in thin films (Cont…)
Newtons and Hooke observed and developed the interference
phenomenon due to multiple reflections from the surface of thin
transparent materials.
Everyone is familiar with the beautiful colours produced by thin
film of oil on the surface of water and also by the thin film of a
soap bubble.
In the case of thin film interference takes place due to (i) reflected
light and (ii) transmitted light.
• Let us consider a transparent film of thickness t and refractive
index . A ray AB is incident on the upper surface of the film
partly reflected R1 and partly refracted along BC. At C part of it is
reflected along CD and finally emerges out as R2.
• The difference in path between the two rays R1 and R2 can be
calculated. In the Figure, BC = CD =CE, DN⊥R1 and BM⊥CD. The
incident angle is i and the refracted angle is r.
Interference effects by in thin films (Cont…)
Interference effects by in thin films (Cont…)
• The optical path difference is given by
x = (BC+CD)-BN = DE-BN
Here  =
𝑠𝑖𝑛 𝑖
sin 𝑟
=
𝐵𝑁/𝐵𝐷
𝐷𝑀/𝐵𝐷
=
𝐵𝑁
𝐷𝑀
 BN = 𝐷𝑀
Hence, x = (DE-DM) = ME
In the right BME
𝑐𝑜𝑠 𝑟 =
𝑀𝐸
𝐵𝐸
 ME = 𝐵𝐸𝑐𝑜𝑠 𝑟 = 𝐵𝐹 + 𝐹𝐸 𝑐𝑜𝑠 𝑟 = 2𝑡𝑐𝑜𝑠 𝑟
So, the optical path difference,
𝑥 = 2𝑡𝑐𝑜𝑠 𝑟
According to EM theory, a ray of light travelling in air and getting
reflected at the surface of a denser medium, undergoes an
automatic phase change of π (or) an additional path difference of
λ/2. Since the reflection at B is at the surface of a denser medium,
there is an additional path difference λ/2 .

The effective path difference in this case, 𝑥𝑡𝑜𝑡𝑎𝑙 = 2𝑡𝑐𝑜𝑠 𝑟 +
2
Interference effects by in thin films (Cont…)
• If the path difference xtotal = m, where m = 0, 1, 2, 3, 4,….. etc.,
constructive interference takes place and the film appears
bright.


 2𝑡𝑐𝑜𝑠 𝑟 + = 𝑚
or, 𝟐𝒕𝒄𝒐𝒔 𝒓 = (𝟐𝒎 − 𝟏)
2
𝟐

• If the path difference 𝑥𝑡𝑜𝑡𝑎𝑙 = (2𝑚 + 1) , where m = 0, 1, 2, 3,
2
4, ….. etc., destructive interference takes place and the film
appears darks.


 2𝑡𝑐𝑜𝑠 𝑟 + = (2𝑚 + 1)
or, 𝟐𝒕𝒄𝒐𝒔 𝒓 = 𝒎
2
2
• If light is incident normally i = 0 and hence r = 0. Therefore the

condition for bright fringe is 2𝑡 = (2𝑚 − 1) and for dark
2
fringe is 2𝑡 = 𝑚.
• For normal incidence, about 4% of the incident light is reflected
and 96% is transmitted. There is a small difference in the
amplitudes of R1 and R2.
Interference effects by in Newton’s ring
• Isaac Newton demonstrated in 1704.
• When a plano-convex lens of long focal length is placed on a
plane glass plate, a thin film of air is enclosed between the
lower surfaces of the lens and the upper surface of the plate.
• The thickness of the air film is very small at the point of contact
and gradually increases from the centre onwards.
Interference effects by in Newton’s ring (Cont…)
• The fringes produced with monochromatic light are circular and
concentric with the point of contact as the centre.
• Near the centre, circular fringe width is higher. As the air film
thickness increases, the width of the bright or dark fringes
decreases.
• When viewed with white light, the fringes are coloured. With
monochromatic light, bright and dark fringes are produced in
the air film.
Interference effects by in Newton’s ring (Cont…)
• In the picture,
S = Monochromatic light source of wavelength  allows to
pass through the lens L1.
B = Glass plate placed at 45 reflects a part of the incident
light towards the air film enclosed by the lens L and the glass
plate G.
M = Microscope which is used to view the dark and bright
interference pattern formed in the air film.
• The centre of the newton’s ring is dark, as there is not path
difference between the interfering rays.
R = radius of curvature of the plano-convex lens.
r = radius of the newton’s rings
For air, refractive index  = 1
According to EM theory (Stokes theory), a ray of light travelling in air and
getting reflected at the surface of a denser medium, undergoes an automatic
phase change of π (or) an additional path difference of λ/2.
The point B is backed by a rarer medium (air) while the point D is backed by a
denser medium (glass). Thus there will be an additional path difference
λ/2 between the rays BC and BDEF corresponding to this additional phase
difference of π.

The effective path difference in this case, 𝑥𝑡𝑜𝑡𝑎𝑙 = 2𝑡𝑐𝑜𝑠 𝑟 ±
2
The two rays will interfere constructively when


2𝑡𝑐𝑜𝑠 𝑟 − = 𝑛
or, 𝟐𝒕𝒄𝒐𝒔 𝒓 = 𝟐𝒏 + 𝟏 with n = 1, 2, 3, …..
2
𝟐
The rays will interfere destructively when


 2𝑡𝑐𝑜𝑠 𝑟 − = (2𝑛 − 1)
or, 𝟐𝒕𝒄𝒐𝒔 𝒓 = 𝒏 with n = 1, 2, 3, …..
2
2
The experimental arrangement is so designed that the light incident at i = 0
and hence r = 0.

Therefore the condition for bright fringe is 2𝑡 = (2𝑛 + 1) and for dark
2
fringe is 2𝑡 = 𝑛.
Interference effects by in Newton’s ring (Cont…)
• The radius of the nth bright fringe ring,
𝑟=
• The radius of the
mth
(2𝑛−1)𝜆𝑅
2𝜇
=
(2𝑛−1)𝜆𝑅
2
dark fringe ring, 𝑟 =
𝑛𝜆𝑅
𝜇
= 𝑛𝜆𝑅
• When n = 1, the radius of the dark fringe is zero and the
radius of the bright
𝜆𝑅
.
2
Therefore, the centre is dark.
• The centre of the ring dark in Newton’s Rings experiment
with reflected light is dark because at the point of contact the
path difference is zero but one of the interfering ray is
reflected so the effective path difference becomes λ/2 thus
the condition of minimum intensity is created hence centre
of ring pattern is dark.
Interference effects by in Newton’s ring (Cont…)
• Alternately dark and bright rings are produced.
• In terms of diameter of the newton’s ring, the wavelength of
the light can be determined from
2
𝐷𝑛+𝑝
− 𝐷𝑛2
𝜆=
4𝑝𝑅
Where Dn+p and Dn are the diameter of the (n+p)th and nth rings.
Michelson Interferometer
FTIR
Fourier Transform Infrared Spectroscopy
LIGO
Laser Interferometer Gravitational Wave
Observatory
2017 Nobel Prize in Physics to Rainer Weiss,
Barry C. Barish, and Kip S. Thorne. Weiss for the
detection of Gravitational Waves: A concept
Given by Albert Einstein
Dipankar Talukdar and Selim Shahriar
were in the LIGO team
Mathematical Problems
1) A viewing screen is separated from a double-slit source by 1.2 m. The distance between the
two slits is 0.030 mm. The second-order bright fringe (m = 2) is 4.5 cm from the center line. (a)
Determine the wavelength of the light. (b) Calculate the distance between adjacent bright fringes.
2) A light source emits visible light of two wavelengths:  = 430 nm and  = 510 nm. The source
is used in a double-slit interference experiment in which L = 1.50 m and d = 0.025 0 mm. (b)
Find the separation distance between the third-order bright fringes. (b) What if we examine the
entire interference pattern due to the two wavelengths and look for overlapping fringes? Are there
any locations on the screen where the bright fringes from the two wavelengths overlap exactly?
3) Suppose you pass light from a He-Ne laser through two slits separated by 0.0100 mm and find
that the third bright line on a screen is formed at an angle of 10.95º relative to the incident beam.
(a) What is the wavelength of the light? (b) Interference patterns do not have an infinite number
of lines, since there is a limit to how big m can be. What is the highest-order constructive
interference possible with the system described in (a)?
4) Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on
double slits separated by 0.100 mm.
5) What is the separation between two slits for which 610-nm orange light has its first maximum
at an angle of 30.0°?
6) Find the distance between two slits that produces the first minimum for 410-nm violet light at
an angle of 45.0°.
7) What is the highest-order maximum for 400-nm light falling on double slits separated by 25.0
μm?
8) (a) If the first-order maximum for monochromatic light falling on a double slit is at an angle
of 10.0°, at what angle is the second-order maximum? (b) What is the angle of the first
minimum? (c) What is the highest-order maximum possible here?
9) (a) Find the thickness of a soap film (µ = 1.33) for a strong first-order reflection of yellow
light,  = 600 nm (in vacuum). Assume normal incidence. (b) What is the wavelength of the light
in the film?
10) A glass plate 0.40 micron thick is illuminated by a beam of white light normal to the plate.
The index of refraction of the glass is 1.50. What wavelength within the limits of the visible
spectrum (= 4010-6 cm to  = 70 10-6 cm) will be intensified in the reflected beam?
11) Newton’s rings are observed when plano-convex is placed convex side down on a plane glass
surface and the system is illuminated from above by monochromatic light. The radius of the first
bright ring is 1 mm. (a) If the radius of the convex surface is 4 m, what is the wavelength of the
light used? (b) If the space between the lens and the flat surface is filled with water, what is the
radius of the first bright ring?