Interference of sound waves - class-XI

The beat frequency is the difference between the frequencies of two waves. The time interval between successive maximum intensities (beats) is the reciprocal of the beat frequency: T = 1/10 = 0.1 s.

If intensity is reduced by 10 percent, 90 percent (0.9) remains. After two slabs, the remaining intensity is 0.9 * 0.9 = 0.81. The total reduction is 1 - 0.81 = 0.19, which is 19 percent.

Intensity is proportional to the square of the amplitude and the square of the frequency (I proportional to A^2 * f^2). Wave 1 has amplitude 2a and frequency f, while Wave 2 has amplitude a and frequency 2f. Thus, intensities are proportional to (2a)^2 * f^2 = 4a^2f^2 and a^2 * (2f)^2 = 4a^2f^2, making them equal. Statement 2 is false because intensity depends on both amplitude and frequency.

Intensity of waves of given frequency in same medium is not only proportional to square of amplitude. It depends on other factors also

Two wave sources are said to be coherent if the waves either have same phase or constant phase difference at any two points on a wave and the phenomenon is known as spatial coherence.

Intensity of a wave is directly proportional to the square of amplitude of the wave.
i.e.  $I \propto A^2$
More is the amplitude of wave, more will be the intensity.
So, we write  $I = kA^2$
where $k$ is a proportionality constant.

The resultant of two waves is given by $R^2=A^2+B^2+2AB cos \theta; \theta$ is the phase difference and A and B are the amplitudes of the individual waves.

Since both the waves are identical, their amplitudes are equal

Thus, $R^2 = A^2+A^2+A^2=3A^2 \implies R=\sqrt{3}A$

Since intensity is proportional to $\sqrt{A}$, we can write the resultant amplitude $R = \sqrt{3I}$

The correct option is (b)

If $I _1$ and $I _2$ are intensities of two waves, then $I _1+I _2=5I$ and $(I _1-I _2)=2I$. Solving, we get,$I _1=3.5I$ and $I _2=1.5I$

The correct option is (b)

$y _{res}$ at point $P=y _1+y _2\=a\cos \omega t+a\cos\omega t\=2\cos\omega t$

The resultant intensity of two waves with equal amplitude 'a' and phase difference 'delta' is I = I1 + I2 + 2sqrt(I1I2)cos(delta). With I1 = I2 = ka^2, I = 2ka^2(1 + cos(delta)) = 4ka^2cos^2(delta/2). Assuming k=1, the result is 4a^2*cos^2(delta/2).

Initial intensity I0 = (sqrt(I1) + sqrt(I2))^2. Since S1 and S2 are identical, I1 = I2 = I, so I0 = (2*sqrt(I))^2 = 4I. Power of S1 reduced by 64 percent means new I1 = 0.36I. New Imax = (sqrt(0.36I) + sqrt(I))^2 = (0.6 + 1)^2 * I = 2.56I. Since I0 = 4I, I = I0/4. Thus Imax = 2.56 * (I0/4) = 0.64I0.

$\begin{array}{l} \Delta x=5\times { 10^{ -9 } }m \ \Delta \phi =\frac { { 2\pi  } }{ \lambda  } \Delta x \ =\frac { { 2\pi \times 5\times { { 10 }^{ -9 } } } }{ { 100\times { { 10 }^{ -10 } } } }  \ =\pi  \ Now, \ I={ I _{ 0 } }+4{ I _{ 0 } }+2\sqrt { { I _{ 0 } } } \sqrt { 4{ I _{ 0 } } } \cos  \phi  \ =5{ I _{ 0 } }+4{ I _{ 0 } }\left[ { \cos  \pi  } \right]  \ ={ I _{ 0 } } \ \therefore resula\tan  t, , , { { intensity } }, ={ I _{ 0 } } \ Hence,, option, , B, , is, the, correct, answer. \end{array}$

A laser beam can be focused on an area equal to the square of its wavelength.

Intensity I = 2 * pi^2 * f^2 * A^2 * rho * v. Plugging in values: I = 2 * (3.14)^2 * (2000)^2 * (10^-8)^2 * 1.3 * 340. I = 2 * 9.86 * 4 * 10^6 * 10^-16 * 1.3 * 340 = 5.3 * 10^-4 W/m^2.

Amplitude of the resultant wave is:
$A _R=\sqrt{A^2 _1+A^2 _2+2A _1A _2\cos(\theta)}$.

Velocity v = distance/time = 600m / 2s = 300 m/s. Wavelength lambda = v/f = 300/500 = 0.6 m. Number of wavelengths = distance / lambda = 600 / 0.6 = 1000.

Interference is observed between waves of the same frequency. Waves 1 (sin wt) and 3 (cos wt = sin(wt + pi/2)) have the same frequency (omega). Wave 2 also has frequency omega, but 1 and 3 are the standard pair for demonstrating interference.

Let, $I _1=I _0  $ and $  I _2=4I _0 $

$\dfrac{I _{max}}{I _{min}}=\dfrac{(\sqrt{I _1}+\sqrt{I _2})^2}{(\sqrt{I _1}-\sqrt{I _2})^2}=49$

$I _{max}=(\sqrt{I _1}+\sqrt{I _2})^2$
$I _{min}=(\sqrt{I _1}-\sqrt{I _2})^2$
Contrast is maximum, when $I _{min}=0$ ie. $I _1=I _2$

The relationship between phase difference (delta phi) and path difference (delta x) is delta phi = (2 * pi / lambda) * delta x. Given delta phi = 60 degrees = pi/3 radians, pi/3 = (2 * pi / lambda) * delta x. Solving for delta x gives delta x = lambda / 6.

Ratio of amplitudes $\sqrt{\dfrac{4}{1}}=\dfrac{2}{1}$
$\dfrac{maximum\ amplitude}{minimum\ amplitude}=\dfrac{2+1}{2-1}=\dfrac{3}{1}$

Resultant intensity of two periodic waves is given by
$I={ I } _{ 1 }+{ I } _{ 2 }+2\sqrt { { I } _{ 1 }{ I } _{ 2 }\cos { \delta  }  } $
where $\delta$ is the phase difference between the waves.
For maximum intensity,
$\delta =2n\pi ;n=0,1,2,...$etc.
Therefore, for zero order maxima, $\cos { \delta  } =1$
${ I } _{ max }={ I } _{ 1 }+{ I } _{ 2 }+2\sqrt { { I } _{ 1 }{ I } _{ 2 } } ={ \left( \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }$
For minimum intensity,
$\delta =\left( 2n-1 \right) \pi ;n=1,2,...$etc.
Therefore, for Ist order minima, $\cos { \delta  } =-1$
${ I } _{ min }={ I } _{ 1 }+{ I } _{ 2 }-2\sqrt { { I } _{ 1 }{ I } _{ 2 } } $
$={ \left( \sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }$
Therefore,  ${ I } _{ max }+{ I } _{ min }={ \left( \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }+{ \left( \sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  \right)  }^{ 2 }$
$=2\left( { I } _{ 1 }+{ I } _{ 2 } \right) $

The correct answer is option(B).

According to the principle of superposition, the resultant wave is
$y = asin(kx - \omega t) + asin(kx + \omega t)$
$= 2a\ sin\ \omega t\ cos\ x$                                                 .....(i)

Let the intensities of the two waves be $I _1$ and $I _2$.
Given :  $I _1:I _2 = 16:1$
Ratio of maximum and minimum intensities  $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{I _1}  +\sqrt{I _2}}{\sqrt{I _1} - \sqrt{I _2}}\bigg)^2$
Or   $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{\frac{I _1}{I _2}}  +1}{\sqrt{\frac{I _1}{I _2}} - 1}\bigg)^2$

Or  $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{16}  +1}{\sqrt{16} - 1}\bigg)^2 = \bigg(\dfrac{4+1}{4-1}\bigg)^2$
$\implies  \ $  $\dfrac{I _{max}}{I _{min}} = \dfrac{25}{9}$

As all the waves are in phase and having same amplitude and frequency

Let the intensities of the two waves be $I _1$ and $I _2$.
Given :  $I _1:I _2 = 9:1$
Ratio
of maximum and minimum intensities  $\dfrac{I _{max}}{I _{min}} =
\bigg(\dfrac{\sqrt{I _1}  +\sqrt{I _2}}{\sqrt{I _1} - \sqrt{I _2}}\bigg)^2$
Or   $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{\frac{I _1}{I _2}}  +1}{\sqrt{\frac{I _1}{I _2}} - 1}\bigg)^2$

Or  $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{9}  +1}{\sqrt{9} - 1}\bigg)^2 = \bigg(\dfrac{3+1}{3-1}\bigg)^2$
$\implies  \ $  $\dfrac{I _{max}}{I _{min}} = \dfrac{16}{4} = \dfrac{4}{1}$

When there are two sinusoidal waves in a string, they cause interference of the waves. The principle of superposition is basic to the phenomenon of interference.

Interference is a fundamental property of all wave phenomena, including both light (electromagnetic waves) and sound (mechanical waves).