Prof. Williams
3.23.2020
Single and Double Slit Interference
Purpose: The purpose of this experiment was to demonstrate the interference of light in using
single slit and double slit. It was demonstrated that how would be the interference pattern when
two different slits were used. The slit width and distance were also calculated from the experiment
and the results were compared with actual values.
III. DATA
Slit# 2
a(Slit Width)=
0.04 mm
d(SlitDistance)=
0.25 mm
=
650 nm
L=
253 cm
Distance Between Bright Spots (inner curves) (cm)
1
2
3
4
Avg
0.64
0.63
0.65
0.64
0.640
Distance Between Dark Spots (inner curves) (cm)
1
2
3
4
Avg
0.63
0.65
0.64
0.63
0.638
Distance Between Bright Spots (outer curves) (cm)
5
6.4
6
4.4
4.2
Distance Between Dark Spots(outer
curves)(cm)
1
2
3
4
4.2
4.3
4.2
4.1
Slit# 1
a(Slit Width)=
0.04 mm
=
650 nm
L=
93.2 cm
Distance Between Bright Spots (cm)
2
3
4
Avg
1.4
1.4
1.4
1.4
1.4
For distant screen assumption tansin=
Distance Between Dark Spots (cm)
ave at slit so
2
3
4
Avg
0.3
0.4
0.4
0.5
0.40
For D>>d this approaches a right angle and= a=slit width
Condition for maximum dsin=m mAD y d
V.
CALCULATIONS
Double Slit:
For the angle that slit slide:
tan0=y/L
Angle from the reference to the first bright point (inner curves)
tan0=y/L=0.63/253
0=arctan(0.63/253)=0.1427
Angle from the reference to the first very dark point (outer curves)
tan0=y/L=4.2/253
0=arctan(4.2/253)=0.9511
For the distance between slits, since the inner curve interference occurred to the double slit, in
the other word distance between slits, those distances must be considered for split distance
calculation.
Intensity
dsinO=mX
Double slit finite width
O(radians)
y= m2L/d
d=m^L/y
m=1
2.7
z.0I*t9'0/z-0I*ESZ*6-0I*OS9*I=p
= 0.00025695m=0.256 mm
%(e) I"##$%&.0&"*+ I 100= I,.-./(,--. )&"*+
100
2.4%. It is perfectly close to the actual value.
For the slit width, since the outer curve interference occurred to the width of the slit, in the other
so that distances must be considered for slit width
asin0=mX
y= mXL/a
a=m^L/