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JEE Main Physics Class 12 Wave Optics Part 1 Questions
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© examsnet.com
Question : 24
Total: 100
In a Young’s double slit experiment, the separation between the slits is 0.15 mm. In the experiment, a source of light of wavelength 589 nm is used and the interference pattern is observed on a screen kept 1.5 m away. The separation between the successive bright fringes on the screen is:
[7 Jan 2020 II]
6.9 mm
3.9 mm
5.9 mm
4.9 mm
Validate
Solution:
Given, distance between screen and slits,
D
=
1.5
m
Seperation between slits,
d
=
0.15
m
m
Wavelength of source of light,
λ
=
589
n
m
Fringe-width,
w
=
D
d
λ
=
1.5
0.15
×
10
−
3
×
589
×
10
−
9
m
=
589
×
10
−
2
m
m
=
5.89
m
m
≈
5.9
m
m
© examsnet.com
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