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JEE Mains 06-Sep-2020 Shift 1 Solved Paper
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© examsnet.com
Question : 34
Total: 75
In the figure below,
P
and
Q
are two equally intense coherent sources emitting radiation of wavelength
20
m
.
The separation between
P
and
Q
is
5
m
and the phase of
P
is ahead of that of
Q
by
90
°
.
A
,
B
and
C
are three distinct points of observation, each equidistant from the midpoint of
PQ
. The intensities of radiation at
A
,
B
,
C
will be in the ratio:
0
:
1
:
2
4
:
1
:
0
0
:
1
:
4
2
:
1
:
0
Validate
Solution:
For (A)
x
P
−
x
Q
=
(
d
+
2.5
)
−
(
d
−
2.5
)
=
5
m
Δ
ϕ
due to path difference
=
2
π
λ
(
Δ
x
)
=
2
π
20
(
5
)
=
π
2
At
A
,
Q
is ahead of
P
by path, as wave emitted by Q reaches before wave emitted by P.
Total phase difference at
A
=
π
2
−
π
2
(due to
P
being ahead of
Q
by
90
°
)
=
0
I
A
=
I
1
+
I
2
+
2
√
I
1
√
I
2
cos
Δ
ϕ
=
I
+
I
+
2
√
I
√
I
cos
(
0
)
=
4
I
For C
x
Q
−
x
P
=
5
m
Δ
ϕ
due to path difference
=
2
π
λ
(
Δ
x
)
=
2
π
20
(
5
)
=
π
2
Total phase difference at
C
=
π
2
+
π
2
=
π
I
net
=
I
1
+
I
2
+
2
√
I
1
√
I
2
cos
Δ
ϕ
=
I
+
I
+
2
√
I
√
I
cos
(
π
)
=
0
For B
x
P
−
x
Q
=
0
Δ
ϕ
=
π
2
(Due to
P
being ahead of
Q
by
90
°
)
I
B
=
I
+
I
+
2
√
I
√
I
cos
π
2
=
2
I
I
A
:
I
B
:
I
C
=
4
I
:
2
I
:
0
=
2
:
1
:
0
© examsnet.com
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