JEE MAIN - Physics (2020 - 5th September Evening Slot - No. 4)
Two coherent sources of sound, S1 and S2,
produce sound waves of the same wavelength,
$$\lambda $$ = 1 m, in phase. S1 and S2 are placed 1.5 m
apart (see fig). A listener, located at L, directly
in front of S2 finds that the intensity is at a
minimum
when he is 2 m away from S2. The listener moves away from S1, keeping his distance from S2 fixed. The adjacent maximum of intensity is observed when the listener is at a distance d from S1. Then, d is :_5th_September_Evening_Slot_en_4_1.png)
when he is 2 m away from S2. The listener moves away from S1, keeping his distance from S2 fixed. The adjacent maximum of intensity is observed when the listener is at a distance d from S1. Then, d is :
12 m
2 m
3 m
5 m
Explanation
_5th_September_Evening_Slot_en_4_2.png)
Initially S2L = 2 m
S1L = $$\sqrt {{2^2} + {{\left( {1.5} \right)}^2}} $$ = 2.5 m
For minimum at (L)
S1L – S2L = $$\Delta $$x = $$\left( {2n + 1} \right){\lambda \over 2}$$
$$ \Rightarrow $$ 2.5 - 2 = $$\left( {2n + 1} \right){1 \over 2}$$
$$ \Rightarrow $$ 0.5 = $${{\left( {2n + 1} \right)} \over 2}$$
$$ \Rightarrow $$ 1 = 2n + 1
$$ \Rightarrow $$ n = 0 (first minima)
so at P $$ \to $$ first maxima
$$ \therefore $$ S1P – S2P = n$$\lambda $$ = 1$$\lambda $$
$$ \Rightarrow $$ d - 2 = 1
$$ \Rightarrow $$ d = 3 m
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