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JEE Main Physics Class 12 Wave Optics Part 1 Questions
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© examsnet.com
Question : 90
Total: 100
A mixture of light, consisting of wavelength 590 nm and an unknown wavelength, illuminates Young’s double slit and gives rise to two overlapping interference patterns on the screen. The central maximum of both lights coincide. Further, it is observed that the third bright fringe of known light coincides with the 4th bright fringe of the unknown light. From this data, the wavelength of the unknown light is:
[2009]
885.0 nm
442.5 nm
776.8 nm
393.4 nm
Validate
Solution:
Let
λ
be the wavelength of unknown light. Third bright fringe of known light coincides with the 4th bright fringe of the unknown light.
∴
3
λ
1
D
d
=
4
λ
D
d
∴
3
(
590
)
D
d
=
4
λ
D
d
⇒
λ
=
3
4
×
590
=
442.5
n
m
© examsnet.com
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