. in young’s experimental set-up, the slit separation is 2 mm and the distance between the slits and the observation screen is 100 cm. calculate the path difference between the waves arriving at a point 5 cm away from the point where the line dividing the slits touches the screen.
Answers 1
Explanation:
Given that: d=2mm=2×10-3m
l=600nm
=6×10-7m
Imax=0.20
W
m2
,D
=2m
for the point y=0.5cm
We know path difference =x=
yd
D
=0.5×10I-2×2×
10-3
2
=5×10-5m
So the corresponding phase difference is
ϕ=
2πx
λ
=
2π×5×10-6
6×10-7
=
50π
3
=
3×16π+2π
3
→ϕ=
2π
3
so the amplitude of the resulting wave at the point y=0.5 cm is
A=
√
r2+r2+2r2
cos(2π)
3
0
√
r2+r2-r2
=r
Since
I
Imax
-
A2
2r2
[since maximum amplitude -2r]
→
I
0.2
=
A2
4r2
=
r2
4r2
→I=
.2
4
=0.05
W
m2
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Author:
cottondii9
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