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JEE Main Physics Class 12 Wave Optics Part 1 Questions
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© examsnet.com
Question : 92
Total: 100
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
[2007]
3
4
1
√
2
√
3
2
1
2
Validate
Solution:
For path difference of
λ
,
the phase difference is
2
π
For path difference of
λ
6
,
the phase difference is
2
π
×
λ
∕
6
λ
=
π
3
Resultant intensity
I
=
I
1
+
I
2
+
2
√
I
1
√
I
2
cos
π
3
∴
I
=
I
1
+
I
2
+
√
I
1
√
I
2
For two identical source,
I
1
=
I
2
=
I
(say)
then
I
=
3
I
′
I
max
=
(
√
I
1
+
√
I
2
)
2
Maximum resultant intensity,
=
(
√
I
′
+
√
I
′
)
2
=
(
2
√
I
′
)
2
=
4
I
′
∴
I
I
max
=
3
4
© examsnet.com
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