Describe Young experiment of interference of light . obtain the conditions of maxima and minima of light.
Answers
Answer:
When light waves from two illuminated slits is incident on the screen, the path traveled by each light wave is different. This path difference leads to a phase difference in the two light waves. The path difference is different for each point on the screen and hence, intensity is different for all the points. This leads to the formation and bright and dark fringes on the screen.
Consider point P on the screen as shown in the figure.
S
2
P
2
=S
2
F
2
+PF
2
S
2
P=
D
2
+(x+
2
d
)
2
Similarly,
S
1
P=
D
2
+(x−
2
d
)
2
Path difference is given by:
S
2
P−S
1
P=
D
2
+(x+
2
d
)
2
−
D
2
+(x−
2
d
)
2
Using binomial expansion,
S
2
P−S
1
P=D(1+
2
1
(
D
x
+
2D
d
)
2
+...)−D(1+
2
1
(
D
x
−
2D
d
)
2
+....)
Ignoring higher order terms,
Δx=S
2
P−S
1
P≈
D
xd
For constructive interference i.e. bright fringes,
nλ=
D
xd
x
n
=
d
nλD
Fringe width is equal to the distance between two consecutive maxima.
β=x
n
−x
n−1
=
d
nλD
−
d
(n−1)λD
β=
d
λD
(b)
I
min
I
max
=
(a
1
−a
2
)
2
(a 1 +a2 ) 2 = 259
Solving,
a 2a 1 = 14
Ratio of slit widths,
w
2w 1= a 22a
12 =16