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You are inside a room, 15 m from the west wall, which has two doors (similar to slits) and you are centered between the two open doors that are 3 m apart. Someone is blaring a 200 Hz tone outside the door so that it enters the doors as a plane wave. You hear a maximum intensity in your current position. a. As you walk (north) along the direction of the west wall with the doors (but maintain a distance 15 m from the west wall), how far will you walk (in m) to hear a minimum in the sound intensityb. What is the spacing (in m) between maxima in sound intensity along that axis?

Respuesta :

Answer:

The distance of first minimum from central maxima is 4.25 m

The spacing between maxima in sound intensity is 8.5 m.

Explanation:

Given that,

Distance D= 15 m

Frequency = 200 Hz

distance between slits a=3 m

We know that,

The width of fringe,

[tex]\beta=\dfrac{\Lambda D}{a}[/tex]

[tex]\beta=\dfrac{vD}{fa}[/tex]

(a). We need to calculate the distance of first minimum from central maxima

Using formula for distance

[tex]d=\dfrac{\beta}{2}[/tex]

[tex]d=\dfrac{vD}{2fa}[/tex]

Put the value into the formula

[tex]d=\dfrac{340\times15}{2\times200\times3}[/tex]

[tex]d=4.25\ m[/tex]

(b). We need to calculate the spacing between maxima in sound intensity

Using formula for the distance of first maxima

[tex]d'=2d[/tex]

[tex]d'=2\times4.25[/tex]

[tex]d'=8.5\ m[/tex]

Hence, The distance of first minimum from central maxima is 4.25 m

The spacing between maxima in sound intensity is 8.5 m.