in a youngs double slit experiment, the distance betwen the two slits is 0.5 mm and the distance between the screen and the slits 1m. when a light of wavelength 500 nm is incident on the slits, what would be distance between the two second bright fringes?
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12th
Physics
Wave Optics
Young's Double Slit Experiment
In Young's double slit expe...
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Asked on December 27, 2019 by
Chinthana Abhista
In Young's double slit experiment the distance between two slits is 2 mm and screen is at a distance of 120 cm from the plane of slits. The smallest distance from the central maxima where the bright fringe due to light of wavelength 6500 A
o
and 5200 A
o
would coincide?
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ANSWER
n
th
maxima is formed at the distance:
y
n
=
d
nλD
At the locations where the maximas of the two wavelengths coincide,
y
n1,λ1
=y
n2,λ2
d
n
1
λ
1
D
=
d
n
2
λ
2
D
n
2
n
1
=
λ
1
λ
2
=
4
5
⟹n
2
=4k;k∈Z
+
∵n
1
∈Z
+
Smallest distance would correspond to n
2
=4
∴y
4,λ
2
=
2×10
−3
4×6500×10
−10
×1.2
=15.6×10
−4
m
=0.156 cm