WAVE OPTICS

A wavefront is a surface of Huygen’s principle of

constant phase the secondary
wavelets-Each point
on the wavefront is
the source of a
secondary wavelets.

WAVE
OPTICS
MIND
MAP

Interference of light

Diffraction Path difference 𝗈= 𝐱𝐝

𝑫

𝐧𝐃𝜆
Position of nth bright fringe 𝒙 𝒏= (where
𝒅
n=0,1,2,3…)
1. Path difference
(𝟐𝐧−𝟏)𝐃𝜆
𝗈= BP-AP = BE =a sin ɵ Position of nth dark fringe 𝒙 𝒏=
𝟐𝒅
( where
2. Position of minima- n=1,2,3…
Position of nth dark 𝝀𝑫
Fringe width 𝛽 = 𝒙 𝒏 − 𝒙 𝒏−𝟏 = 𝖰𝒃𝒓𝒊𝒈𝒉𝒕 = 𝖰𝒅𝒂𝒓𝒌 =
fringe 𝒅

a sin ɵn = nλ Fringe width in medium 𝖰𝒎𝒆𝒅 =

𝖰𝒗𝒂𝒄

(where n =1,2,3… ) 𝝁

3.Position of secondary Angular width of fringe 𝜽 =

𝖰
=
𝝀
𝑫 𝒅
maxima-
a sin ɵn’ = (2n+1) λ/2
where n =1,2,3…
4.angular width of
central maxima = 2ɵ
=2λ/a
5. Linear width of
central maxima= 2x or
D(2ɵ) = 2λD/a
GIST OF THE LESSON
1. Nature of light-The phenomena like interference, diffraction and polarization
establish the wave nature of light. Whereas the phenomena like photo electric effect,
Raman effect, Compton effect establish the particle nature of light.

2. Wavefront-It is defined as the continuous locus of all the particles of the medium
vibrating in the same phase at any instant. A wavefront is a surface of constant phase. The
speed with which the wavefront moves outwards from the source is called the phase speed
(wave speed).
Note-1. Rays are perpendicular to wavefronts. .2. No backward wavefront is possible.

3. Types of wavefront-It is depends on the source of disturbance.

Spherical Wavefront formed by the point

wavefront
source

Cylindrical Wavefront formed by linear or

wavefront
cylindrical shape source

As a spherical or cylindrical
Plane wavefront advances, its curvature
wavefront
decreases, so small portion of such a
wavefront at a large distance from
the source will be a plane wavefront

4. Huygen’s principle of
the secondary Wavelets-
It is the basis of wave theory of
light. It tells how a wavefront
propagates through a medium.
It is based on the following
assumptions
i) Each point on a wavefront
acts as a source of new disturbance called secondary wavelets. These secondary wavelets
spread out in all directions with the speed of light in the given medium.
ii) The wavefront at any later time is given by the forward envelope of the secondary
wavelets at that time.

5. During refraction- Frequency of light remains constant, wavelength and speed of

light get changed depending on the refractive index. (λ’= λ/μ and v’= v/μ) (here μ is the
refractive index)

6. Behaviour of a prism, lens and mirror-

7.
Reflection on the basis of wave theory Refraction on the basis of wave theory
i)

In triangle 𝗈ABC and 𝗈DCB From 𝗈ABC, sin i =BC/AC

∠BAC=∠CDB (Each 90o) From 𝗈ADC, sin r = AD/AC
sin i 𝐵𝐶 𝑣1𝑡
BC=BC ∴ sin = =
𝑟 𝐴𝐷 𝑣2 𝑡
AC=BD (each equal to ct) Or sin i =
𝑣1
= 𝜇 (refractive index of
ii) ∴𝗈ABC ≌ 𝗈DCB sin 𝑟 𝑣2 21
Hence ∠i=∠r second medium wrt first medium)

Note-for denser to rarer medium

8. Coherent and Incoherent Sources-Two sources are coherent if they have the
same frequency and with a constant phase difference. They are incoherent if phase
difference is not constant.

9.Interference of light-When two light waves of the same frequency and having
constant phase difference(coherent), travelling in the same direction superpose each other,
the intensity gets redistributed, becoming maximum at some points and minimum at
others, this phenomenon is called interference of light.
Let two waves from two coherent source of light be 𝑦1 = 𝑎 cos 𝜔𝑡 and 𝑦2 = a cos(𝜔𝑡 + Ø)
Where a and b are amplitudes and Ø is the phase difference

y = R cos(𝜔t + Ø/2 )

R = 2 a c o s Ø/2

Inet = 4I0cos2 (Ø/2)

Note:- If amplitudes are not same, then
Let two waves from two coherent source of light be 𝑦1 = 𝑎 sin 𝜔𝑡 and 𝑦2 = 𝑏 sin(𝜔𝑡 + Ø)
Where a and b are amplitudes and Ø is the phase difference
So y = y1 + y2 after solving we get y = Asin(𝜔t + 𝜃)

 Where A is the resultant amplitude so Anet = √(a2 + b2 + 2abcosØ

 And Resultant intensity is Inet = I1 + I2 + 2√I1I2 cosØ

Ø
 Resultant amplitude when a = b 𝐴𝑛𝑒𝑡 = 2𝑎 𝑐𝑜𝑠
2

Ø
 Resultant intensity when I1=I2=I 𝐼𝑛𝑒𝑡 = 4𝐼 𝑐𝑜𝑠2
2

NOTE- Ratio of maximum intensity to minimum intensity

2
𝐼𝑚𝑎𝗑 𝑎+𝑏 2 √𝐼1 + √𝐼2
=( ) =( )
𝐼𝑚i𝑛 𝑎−𝑏 √𝐼1 + √𝐼2

10. Types of Interference-

s.no Constructive interference Destructive interference
1 Point where resultant intensity is max Point where resultant intensity is minimum
 2

3 Resultant intensity at a point is Resultant intensity at a point is

maximum when the phase difference minimum when the phase difference is
is even multiple of π or path odd multiple of π or path difference is
difference is an integral multiple of an odd multiple of wavelength λ/2
wavelength λ

11. Young’s Double Slit Experiment-It is the practical verification of interference.

In this we get two coherent sources by dividing wavefront. We always get bright fringe at
the center of the screen and both side alternately bright and dark fringes are made.
a) Fringe width in YDSE-
In 𝗈S1S2L
𝑆2𝐿 𝗈
sin = =
𝑆1𝑆2 𝑑
X
Now in 𝗈DOP tan =
𝐷

If ɵ is small sin ɵ ≈tan ɵ ≈ ɵ

X
So 𝗈 =
𝑑 𝐷
xd
b)Path difference 𝗈=
𝐷
nD𝜆
c) Position of nth bright fringe 𝗑𝑛 = where n=0,1,2,3…
𝑑
(2n−1)D𝜆
d) Position of nth dark fringe 𝗑𝑛 = where n=1,2,3…
2𝑑

e) Fringe width –Separation between position two consecutive maxima or minima. Width
of bright and dark fringe will be same.
 𝐷
𝛽 = 𝗑𝑛 − 𝗑𝑛−1 = 𝖰𝑏𝑟i𝑔ℎ𝑡 = 𝖰𝑑𝑎𝑟𝑘 =
𝑑
𝖰𝑣𝑎𝑐
f)Fringe width in medium 𝖰𝑚𝑒𝑑 = 𝜇
𝖰
g) Angular width of fringe 𝜃 = =
𝐷 𝑑

h) Overlapping of fringes
if n1thbright fringe overlapped on n2th bright fringe then n1λ1= n2λ2
if bright overlapped dark n1λ1= (2n2 -1) λ2/2
𝐷
i)Dependence of fringe width 𝛽= (β α λ, β α D, β α 1/d)
𝑑
j) Intensity distribution curve-

k) Condition for sustained interference-

i) Two source of light must be coherent(ii) Having same frequency (iii)source should be
monochromatic (iv)wave must travel in same direction(v) for a better contrast amplitude
of waves should be approximately equal

12. Diffraction

It is the phenomena of bending of light around corners of an obstacle or aperture in the

path of light. Due to this bending, light goes into the geometrical shadow region of the
obstacle or aperture.
This bending becomes more when the dimensions of the aperture or the obstacle are
comparable of the wavelength of light.

13. Diffraction of light from a single slit-

a) Central maxima-maximum intensity at point o because path difference at o is zero.

b) Path difference 𝗈= BP-AP = BE =a sin ɵ

c) Position of minima- Position of nth dark fringe a sin ɵn = nλ where n =1,2,3…

d) Position of secondary maxima- a sin ɵn’ = (2n+1) λ/2 where n =1,2,3…

e) Width of central maxima- the direction of first minima ɵ=λ/a, this angle is called half
angular width of central maxima
Note-width of secondary
angular width of central maxima = 2ɵ =2λ/a 1
maxima α
𝑠𝑙𝑖𝑡 𝑤𝑖𝑑𝑡ℎ
f) Linear width of central maxima= 2x or D(2ɵ) = 2λD/a
g) Graph

Wave Optics Important Questions (with answers)

(1) When a low flying air craft passes over the head we sometimes notice slight shaking of
the picture on our TV screen. Explain.
Ans: Interference of the direct signal received by the antenna with the signal reflected by
the aircraft.

(2) A diffraction pattern is obtained using a beam of red light. What happens if the red light
is replaced by blue light?
Ans: diffraction pattern becomes narrower and crowded together.

(3) In Young’s experiment, the third bright band for wavelength of light 6000 Å coincides
with the fourth bright band for another source in the same arrangement. The wave length
of the another source is (Ans: 4500 Å)

(4) In Young’s double slit experiment, the intensity ratio of two coherent sources are 81:1.
Calculate the ratio between maximum and minimum intensities. (25:16)

(5) A monochromatic light of wavelength 600nm is incident on a water surface having

refractive index 4/3. Find the velocity, frequency and wavelength of light in water.
(2.25 × 108 m s–1, 5x1014Hz, 450nm)
(6) In Young’s experiment a light of frequency 6 × 1014 Hz is used. Distance between the
centers of adjacent fringes is 0.75 mm. Calculate the distance between the slits, if the screen
is 1.5 m away. (1 mm)

(7) The fringe width obtained in Young’s double slit experiment while using a light of
wavelength 5000 Å is 0.6 cm. If the distance between the slit and the screen is halved, find
the new fringe width. (3 mm)

(8) What is the effect on the interference fringes in Young’s double slit experiment if,
(a) The screen is moved away from the plane of the slits.
(b) Source is replaced by another source of shorter wavelength.
Ans (a) Angular separation of the fringes remains constant and linear fringe width
increases.
(b) Separation of the fringes decreases

(9) During the refraction of wave which physical quantity of wave remains unchanged out
of speed, frequency and wavelength?
Ans. The frequency of the wave remains unchanged, whereas velocity and wavelength
changes.
(10) What will be the effect on phase in refraction and reflection?
Ans: In refraction of wave no change of phase occurs, whereas in reflection from denser
medium there is change of phase π, thus incident and reflected waves have phase
difference of π.
(11) How is wave front related to the direction of propagation of the waves?
Ans: Wave front is always perpendicular to the direction of propagation of the wave
(12) Explain the statement' light falling on light can produce darkness'
Ans: This is possible when light from two coherent sources having equal amplitude
superpose on each other and produces destructive interference.
(13) In YDS experiment the intensity of central maximum is I, what will be the intensity if
one of the slits is closed?
Ans: If light from both the slits reaches at the center of the screen, the intensity is I. If one
slit is closed no interference of light takes place, thus amplitude of resultant wave is halved
and intensity becomes one fourth.
(14) What happens to the energy at the points of destructive interference?
Ans: The energy which disappears at the points of destructive interference will reappear at
the points of constructive interference. Thus, interference is the phenomenon of
redistribution of energy and energy remains conserved.
(15) What will be effect on the fringe width if entire YDSE set up is immersed in water?
Ans: As β = Dλ/d. In water D and d remains unchanged whereas λ decreases therefore β
also decreases.
(16) Why is interference not detected when the two coherent sources are far apart?
Ans: In interference the fringe width is inversely proportional to the distance between the
two coherent sources of light. Thus, if the sources are far apart to each other the fringe
width may become so small that it becomes invisible.
(17) Light from two coherent sources reaches a point where the path difference for yellow
light is 3λ/2. What type of fringe is obtained at that point?
Ans: If monochromatic yellow light is used, then path difference of 3λ/2 results in
destructive interference and it will be dark fringe
(18) In YDS experiment two independent bulbs of same power and material are used. What
will be observed on the screen?
Ans: Two independent sources of light can never be coherent as the phase difference
between them changes quickly and randomly, thus the position of maxima and minima also
changes. Thus, there will be general illumination of the screen without interference.
(19) Which phenomenon establishes the wave nature of light?
Ans: Interference and diffraction
(20) In Young's double slit blue, red and green light are used separately. In which case
will be the fringe width maximum?
Ans: Fringe width β = λD/d, thus the fringe width is directly proportional to λ as D and d
are constant. Thus, it will be maximum for red as its wavelength is maximum.

Unsolved question
One mark questions
(1) What type of wave front will emerge from a (i) point source (ii) distant light source
(2) Give the phase difference between two points on a wave front.
(3) Write the necessary condition for sustained interference of light.
(4) Why is diffraction of sound waves more easily observed than diffraction of light waves?
(5) Obtain the ratio of maxima and minima in an interference pattern.
(6) What is principle of superposition of waves?
 Two marks questions
(1) What is the effect in interference pattern in Young’s double slit experiment due to each
of the following operation (i) separation between slit is increased (ii) Whole apparatus is
kept in denser medium?
(2) Draw the wave front for a (i) converging ray (ii) diverging rays from a point source.
(3) Estimate the distance for which ray optics is a good approximation for an aperture of
4nm and wavelength 400nm?
(4) Write at least three differences between interference and diffraction.
(5) Find out an expression for width of central maxima during diffraction of light.

(6) The ratio of the intensities at minima to maxima in the interference pattern is 9:25.
What will be the ratio of the widths of the two slits in the Young’s double slit experiment?

Three marks questions

(1) Define the term ‘wave front’. Using Huygens’s construction of a wave front, explain the
refraction of a plane wave front at a plane surface and hence deduce Snell’s law.
(2) Derive an expression for fringe width of interference fringes in Young’s double slit
experiment. Show that bright and dark fringes are having equal widths.
(3)(a) Consider a point at the focal point of a convergent lens. Another convergent lens of
short focal length is placed on the other side. What is the nature of the wave fronts
emerging from the final image?
(b)What is the shape of the wave front on earth for sunlight?
(c)Is Huygens principle valid for longitudinal sound waves?
(4) What is interference? Draw the Intensity distribution curve during interference of light.
How does interference pattern changes (i) increasing the separation between the slits (ii)
increasing the screen distance and (iii) using white light?

(5) What is diffraction? Write the essential condition for it. Draw the Intensity distribution
curve. Obtain an expression for the width of central maximum. How does diffraction
pattern changes on using white light?