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KCET 2018 Physics Solved Paper
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© examsnet.com
Question : 24
Total: 60
In Young's double slit experiment ,slits are separated by 2 mm and the screen is placed at a distance of 1.2 m from the slits. Light consisting of two wavelengths 6500 Å and 5200 Å are used to obtain interference fringes. Then the separation between the fourth bright fringes of two different patterns produced by the two wavelengths is
0.312 mm
0.123 mm
0.213 mm
0.412 mm
Validate
Solution:
We know that,
y
=
n
λ
D
d
where λ is wavelength; D is distance of screen from slits; d is separation between the slits. So,
y
1
=
n
λ
1
D
d
and
y
2
=
n
λ
2
D
d
Therefore,
y
1
−
y
2
=
n
D
d
(
λ
1
−
λ
2
)
Given,
n
=
4
;
D
=
1.2
m
;
d
=
2
m
m
=
2
×
10
−
3
m
;
λ
1
=
6500
Å
=
6500
×
10
−
10
m
;
λ
2
=
5200
Å
=
5200
×
10
−
10
m
y
1
−
y
2
=
4
×
1.2
2
×
10
−
3
(
6500
×
10
−
10
−
5200
×
10
−
10
)
=
2.4
×
10
3
×
1300
×
10
−
10
m
=
3120
×
10
−
7
=
0.3120
×
10
−
3
m
⇒
y
1
−
y
2
=
0.3120
m
m
Therefore, the separation between the fourth bright fringesof two different patterns produced by the two wavelengths is 0.3120 mm.
© examsnet.com
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