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JEE Main 24 Feb 2021 Shift 1 Solved Paper
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© examsnet.com
Question : 20
Total: 90
In a Young's double slit experiment, the width of the one of the slit is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit-width. Find the ratio of the maximum to the minimum intensity in the interference pattern.
24 Feb 2021 Shift1
4
:
1
2
:
1
1
:
4
3
:
1
Validate
Solution:
Given, amplitude
∝
width of slit
⇒
A
2
=
3
A
1
We know that,
I
max
I
min
=
(
√
I
1
+
√
I
2
)
2
(
√
I
1
−
√
I
2
)
2
∵
Intensity,
l
∝
A
2
∴
I
max
I
min
=
(
A
1
+
A
2
)
2
(
|
A
1
−
A
2
|
)
2
=
(
A
1
+
3
A
1
|
A
1
−
3
A
1
|
)
)
2
=
(
4
A
1
2
A
1
)
2
=
4
1
∴
I
max
:
I
min
=
4
:
1
© examsnet.com
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