Question 123: Wave Optics

Question

Two coherent sources of sound, S
1
and S
2
,
produce sound waves of the same wavelength,
$$\lambda $$ = 1 m, in phase. S
1
and S
2
are placed 1.5 m
apart (see fig). A listener, located at L, directly
in front of S
2
finds that the intensity is at a
minimum
when he is 2 m away from S
2
. The
listener moves away from S
1
, keeping his
distance from S
2
fixed. The adjacent maximum
of intensity is observed when the listener is at
a distance d from S
1
. Then, d is :

Accepted Answer

Initially S
2
L = 2 m
S
1
L = $$\sqrt {{2^2} + {{\left( {1.5} ight)}^2}} $$ = 2.5 m
For minimum at (L)
S
1
L Ã¢ÂÂ S
2
L = $$\Delta $$x = $$\left( {2n + 1} ight){\lambda  \over 2}$$
$$ \Rightarrow $$ 2.5 - 2 = $$\left( {2n + 1} ight){1 \over 2}$$
$$ \Rightarrow $$ 0.5 = $${{\left( {2n + 1} ight)} \over 2}$$
$$ \Rightarrow $$ 1 = 2n + 1
$$ \Rightarrow $$ n = 0 (first minima)
so at P $$ 	o $$ first maxima
$$ 	herefore $$ S
1
P Ã¢ÂÂ S
2
P = n$$\lambda $$ = 1$$\lambda $$
$$ \Rightarrow $$ d - 2 = 1
$$ \Rightarrow $$ d = 3 m

Answer

12 m

Answer

2 m

Answer

3 m

Answer

5 m

Wave Optics

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