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Svetach [21]
3 years ago
11

Two loudspeakers emit sound waves along the x-axis. The sound has maximum intensity when the speakers are 19 cm apart. The sound

intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 29 cm.
What is the wavelength of the sound?

If the distance between the speakers continues to increase, at what separation will the sound intensity again be a maximum?
Express your answer using two significant figures.
Physics
1 answer:
Roman55 [17]3 years ago
6 0

Answer:

a.96cm

b.77cm

Explanation:

Highest intensity,d_1=19cm

Lowest intensity,d_2=29cm

d_1+d_2=19+29=48cm  

48cm is the distance for half-wave, wavelength =48cm*2=96cm

b. From a, above we know that half-wavelength is 48cm

Also, the distance of known  maximum is 29cm

Next maximum is calculated as:

Next \m\aximum=Last \ Maximum+ \ 0.5Wavelength\\=29cm+48cm\\=77cm

The next maximum will occur at 77cm

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Answer:

<em>Part A</em><em>:</em>

a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.

<em>Part B</em><em>:</em>

b) If the spacing between the slits is decreased the fringe spacing Δy will increase.

<em>Part C</em><em>:</em>

a) If the distance to the screen is decreased the fringe spacing will decrease.

<em>Part D</em><em>:</em>

The dot in the center of fringe E is 920\ x\ 10^{-9} m farther from the left slit than from the right slit.

Explanation:

In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.

The position of bright fringes in the screen where the pattern is formed can be calculated with

                      \vartriangle y =\frac{m \lambda L}{d}

                      m=0,\pm 1,\pm 2,\pm 3,.....

  1. m is the order number.
  2. \lambda is the wavelength of the monochromatic light.
  3. L is the distance between the screen and the two slits.
  4. d is the distance between the slits.
  • Part A:  a) In the above equation for the position of bright fringes we can see that if the wavelength of the light \lambda is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
  • Part B:  b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
  • Part C:  a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
  • Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at  the center of the fringe E in the screen we use the condition for constructive interference. That says that the  path length difference Δr between rays coming from the left and right slit must be \vartriangle r=m \lambda

        We simply replace the values in that equation :

                      \vartriangle r= m \lambda =2.\ 460\ nm

                      \vartriangle r= 920\ x\ 10^{-9} m

         The dot in the center of fringe E is 920\ x\ 10^{-9}m farther from the left slit than from the right slit.

     

       

       

     

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