A of Light consisting of wavelength 756
f 600nm is used to obtain interface
fringes in a young's double slith experiment
(Given D = 1m, d = ararn)
(a) Find the distance of the 5th dark fringe
on the screen from the central fringe
for Wonelength 600mm
(6) What's the least distance from the
central fringe where the bright fringes due
to both the wavelengths coincide?
Answers
Fringe width is given by β=
d
λD
In the first case, λ
1
=6000 A
∘
=6000×10
−10
m
β
1
=6 mm=0.006 m
∴0.006=6000×10
−10
d
D
....(1)
In the second case, the distance between the slits and the screen is
2
D
,β
2
=4 mm=0.004 m
∴0.004=λ
2
×
d
2
D
....(2)
Directing equations (2) by (1) we get
0.006
0.004
=
2×6000×10
−10
λ
2
λ
2
=
0.006
0.004×2×6000×10
−10
λ
2
=
0.006
48×10
−10
λ
2
=8000×10
−10
λ
2
=8000 A
∘
.
Answer:
Light of wave length 6000A∘ is used to obtain interference fringes ...
1 answer
Calculate the wave length of light required to obtain fringe of width 4 mm when the distance ... Problems on Young's Double Slit Experiment.
Top answer
Fringe width is given by β = λDd In the first case, λ1 = 6000 A∘ = 6000 × 10-10m β1 = 6 mm = 0.006 m 0.006 = 6000 × 10-10 Dd ....(1) In the second ... More
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