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. (1) 4:1 (2) 2:1 (3) 3:1 (4) 1:4
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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
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