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JEE Main 1-Sep-2021 Shift 2 Solved Paper
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© examsnet.com
Question : 22
Total: 90
The width of one of the two slits in a Young's double slit experiment is three times the other slit. If the amplitude of the light coming from a slit is proportional to the slit-width, the ratio of minimum to maximum intensity in the interference pattern is
x
:
4
where
x
is .........
[1 Sep 2021 Shift 2]
Your Answer:
Validate
Solution:
Given,
b
1
=
3
b
2
Here,
b
1
= width of the one of the two slit
and
b
2
= width of the other slit.
As we know that,
Intensity, I
∝
(
Amplitude
)
2
⇒
I
1
I
2
=
(
b
1
b
2
)
2
⇒
I
1
I
2
=
(
3
b
2
b
2
)
2
As we know, the ratio of the minimum intensity to the maximumintensity in the interference pattern,
I
min
I
max
=
(
√
I
1
−
√
I
2
√
I
1
+
√
I
2
)
2
Substituting the values in the above equations, we get
I
min
I
max
=
(
√
9
I
2
−
√
I
2
)
2
(
√
9
I
2
+
√
I
2
)
2
=
√
(
3
−
1
3
+
1
)
2
I
min
I
max
=
1
4
Comparing with,
I
min
I
max
=
x
4
The value of x = 1.
© examsnet.com
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