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AIEEE 2007 Solved Paper
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© examsnet.com
Question : 72
Total: 118
In a Young's double slit experiment the intensity at a point where the path difference is
λ
6
(
λ
being the wavelength of light used ) is
I
. If
I
0
denotes the maximum intensity,
I
I
0
is equal to
3
4
1
√
2
√
3
2
1
2
Validate
Solution:
The intensity of light at any point of the screen where the phase difference due to light coming from the two slits is
φ
is given by
I
=
I
0
cos
2
(
φ
2
)
where
I
0
is the maximum intensity.
NOTE : This formula is applicable when
I
1
=
I
2
.
Here
φ
=
π
/
3
∴
I
I
0
=
cos
2
π
6
=
(
√
3
2
)
2
=
3
4
© examsnet.com
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