The width of one of the two slits in a Young's double slit experiment is double of the other slit.
Assuming that 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.
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Answer:
Let the amplitude of light wave coming from the narrower slit be a then the amplitude of light wave from the wider slit =2a
The maximum intensity occurs where the constructive interference takes place and the minimum intensity where the destructive interference takes place.
⇒a
max
=2a+a=3a
⇒a
min
=2a−a=a
Ratio of maximum to minimum intensity,
I
min
I
max
=
a
min
2
a
max
2
=
a
2
3
2
a
2
=9
Explanation:
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