1. Optics
Presented to: Dr. Kamran sb.
Presented by: Tahir Younus
Reg. No. BS-PHYS-18-44 (E)
Semester: 2018-22 5th
Department of Physics
Ghazi University Dera Ghazi
Khan
2. β’Introduction
β’Performed by Young in 1801 to demonstrate wave behavior of light.
β’In this experiment, Young identified the phenomenon called interference
β’The double-slit experiment is a demonstration that light and matter can display
characteristics of both waves and particles.
3. Interference occurred during experiment is of two types
ο Constructive Interference
ο Destructive Interference
4. Experiment
ο± A screen having two narrow
slits is illuminated by
monochromatic light.
ο± Superposition of wavelets
result in series of bright and
dark fringes.
ο± The bright fringes are termed
as maxima and dark fringes
are termed as minima.
7. Maxima
If the point P is to have a bright fringe, the path difference must be an integral multiple of π.
BD = mπ
m = 0,1,2,3,4,β¦.
From βπ΄π΅π·.
BD = d sinπ
So, d sinπ= mπ β¦ . . 1)
8. Distance between Adjacent Bright Fringes
Since angle βπβ is very small,
Sin π β ππππ
So, Equation 1 becomes
dππππ=m π β¦ . . 2)
From figure
Tanπ=
π
πΏ
So, equation 2 can be written as
d
π
πΏ
= m π
Y=m
ππΏ
π
β¦ . . 3)
9. If there would be a dark fringe at point P then,
Y=(π +
1
2
)
ππΏ
π
β¦. 4)
Thatβs the distance between two adjacent bright fringes, mth and (m+1) fringes are considered.
For mth bright fringe
π¦π = π
ππΏ
π
For (m+1) bright fringe,
π¦π+1 = (π + 1)
ππΏ
π
Let ββπβ²
be the distance between these two adjacent fringes, then
βY=ππ+1 β ππ= π +
1
2
ππΏ
π
β π
ππΏ
π
βY=
ππΏ
π
10. Minima (Dark fringes)
If the point P is to have a dark fringe, the path difference BD must be half integral number of wavelength
π.
π΅π· = (π +
1
2
)π m=0,1,2,β¦.
From βπ΄π΅π·.
BD = d sinπ
d sinπ = (π +
1
2
)π
First dark fringe is obtained at m=0 and second dark for m=1.
11. Distance between adjacent dark fringes
Since angle βπβ is very small,
Sin π β ππππ
dππππ = (π +
1
2
)π
From figure,
Tanπ=
π
πΏ
d
π
πΏ
= (π +
1
2
)π
If there would be a dark fringe at βpβ then
Y=(π +
1
2
)
ππΏ
π
Now distance between to adjacent dark fringes,
βY=
ππΏ
π
12. Dark and bright fringes are of equal width and are equally spaced.
Fringe spacing is directly proportional to βπβ and distance between slits βLβ.
Inversely to separation of slits βdβ.