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Interference of sound waves - class-XI

Description: Interference of sound waves
Number of Questions: 35
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Tags: superposition of waves waves oscillations and waves superposition of waves-1: interference and beats physics
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If the difference between the frequencies of two waves is 10 Hz then time interval between successive maximum intensity is:

  1. $10 s$

  2. $1 s$

  3. $0.1 s$

  4. $0.01 s$


Correct Option: C

The intensity of the sound gets reduced by $10$% on passing through a slab. The reduction in  intensity on passing through two consecutive slab, would be 

  1. $20$%

  2. $50$%

  3. $19$%

  4. $5$%


Correct Option: C

Statement-1:
Two longitudinal waves given by equations; ${ y } _{ 1 }$(x,t) = 2a $\sin { \left( \omega t-kx \right)  } $ and ${ y } _{ 2 }\left( x,t \right) $ = a $\sin { \left( 2\omega t-2kx \right)  } $  will have equal intensity.

Statement-2:
Intensity of waves of given frequency in same medium is proportional to square of amplitude only.

  1. Statement-1 is true, statement -2 is true; statement-2 is not correct explanation of statement -1.

  2. Statement-1 is false, statement -2 is true.

  3. Statement-1 is true, statement -2 is false.

  4. Statement -1 is true, statement-2 true; statement-2 is the correct explanation of statement-1.


Correct Option: C

Two coherent sources of different intensities send waves which interfere. If the ratio of maximum and minimum intensity in the interference pattern is $25$ then find ratio of intensity of source :

  1. $25 : 1$

  2. $5 : 1$

  3. $9 : 4$

  4. $25 : 16$


Correct Option: C
Explanation:

$\cfrac { { I } _{ max } }{ { I } _{ min } } ={ \left[ \cfrac { \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  }{ \sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  }  \right]  }^{ 2 }$

Where and  are intensities of two waves 

given 

$\cfrac { { I } _{ max } }{ { I } _{ min } } =\cfrac { 25 }{ 1 } \\ \therefore \cfrac { \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  }{ \sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  } =\cfrac { 5 }{ 1 } $

use componendo and dividendo

we get

$\cfrac { \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } } +\sqrt { { I } _{ 1 } } -\sqrt { { I } _{ 2 } }  }{ \sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } } -\sqrt { { I } _{ 1 } } +\sqrt { { I } _{ 2 } }  } =\cfrac { 5+1 }{ 5-1 } \\ or,\quad \cfrac { \sqrt { { I } _{ 1 } }  }{ \sqrt { { I } _{ 2 } }  } =\cfrac { 3 }{ 2 } \\ or,\quad \cfrac { { I } _{ 1 } }{ { I } _{ 2 } } =\cfrac { 9 }{ 4 } $

 

 

Statement -1:
Two longitudinal waves given by equation $y _{1}$(x,t) = 2a sin $(\omega - kx)$ and $y _{2}$(x,t) = a sin $(2\omega - 2kx)$ will have equal intensity.
Statement -2:
Intensity of waves of given frequency in the same medium is proportional to the square of amplitude only.

  1. Statement -1 is true, statement -2 is true; statement -2 is not correct explanation of statement-1.

  2. Statement -1 is false, statement -2 is true.

  3. Statement -1 is true, statement -2 is false

  4. Statement -1 is true, statement -2 true; statement -2 is the correct explanation of statement -1


Correct Option: C
Explanation:

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

If two waves maintain constant phase difference or same phase at any two points on a wave is known as spatial coherence.

  1. True

  2. False


Correct Option: A
Explanation:

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 (I) is related to amplitude (A) as:

  1. $I \propto A$

  2. $I \propto {A^2}$

  3. $I \propto {A^4}$

  4. $I \propto {A^{-1}}$


Correct Option: B
Explanation:

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 intensity for two identical waves of intensity I with a phase difference of $\pi/3$ is 

  1. $R=2\sqrt{3I}$

  2. $R=\sqrt{3I}$

  3. $R=4\sqrt{3I}$

  4. $R=3\sqrt{3I}$


Correct Option: B
Explanation:

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 the sum of the intensities of two component waves are 5I units and upon superposition with a phase difference of $\pi $ radians, their resultant is 2I, what are the intensities of component waves

  1. $I _1=3I, I _2=I$

  2. $I _1=3.5I, I _2=1.5I$

  3. $I _1=2I, I _2=3I$

  4. $I _1=I, I _2=4I$


Correct Option: B
Explanation:

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)

Waves from two sources superpose on each other at a particular point amplitude and frequency of both the waves are equal. The ratio of intensities when both waves reach in the same phase and they reach with the phase difference of $90^{\circ}$ will be

  1. $1 : 1$

  2. $\sqrt {2} : 1$

  3. $2 : 1$

  4. $4 : 1$


Correct Option: C
Explanation:
Relation between intensity $f$
Amplitude
$I= kA^{2} \cos^{2} \left( \dfrac{|theta}{2} \right)$
When both the waves reach in the same phase, then 
$\theta= 0^{o}$ 
$I _{1}= kA^{2} \cos^{2} \left( \dfrac{0^{o}}{2} \right)$
$I _{1} = kA^{2} [\cos 0^{o}=1]$ ------- $(1)$
When both the waves reach with phase difference $90^{o}$.
$\theta = 90^{o}$
$I _{2}= kA^{2} \cos^{2} \left( \dfrac{90^{o}}{2} \right)$
$I _{2} = kA^{2} \cos^{2} (45^{o})$
$I _{2}= kA^{2} \left( \dfrac{1}{2} \right)$ ----- $(2)$
$\dfrac{I _{1}}{I _{2}}= \dfrac{kA^{2}}{kA^{2} (1/2)} =\dfrac{2}{1}$

Suppose displacement produced at some point $P$ by a wave is $y _1=a cos \omega t$ and by another wave is $y _2=a cos \omega t$.Let $I _0$ represents intensity produced by each one of individual wave, then resultant intensity due to overlapping of both wave is

  1. $I _0$

  2. $2I _0$

  3. $\dfrac{I _0}{2}$

  4. $4I _0$


Correct Option: D
Explanation:

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

Now amplitude $=2a$
Since  intensity $\propto$(amplitude)$^2$
So, $\cfrac{I _{new}}{I _0}=\cfrac{4a^2}{a^2}\\implies I _{new}=4I _0$

Two waves $Y _{1}=a\sin \omega t$ and $Y _{2}=a\sin (\omega t+\delta)$ are producing interference, then resultant intensity is-

  1. $a^{2}\cos^{2}\delta/2$

  2. $2a^{2}\cos^{2}\delta/2$

  3. $3a^{2}\cos^{2}\delta/2$

  4. $4a^{2}\cos^{2}\delta/2$


Correct Option: D

Sounds from two identical $S _1$ and $S _2$ reach a point  P. When the sounds reach directly, and in the same phase, the intensity at $P$ is $I _0$. The power of $S _1$ is now reduced by $64\%$ and the phase difference between $S _1$ and $S _2$ is varied continuously. The maximum and minimum  intensities recorded at P are mow $I _{max}$ and $I _{min}$ 

  1. $I _{max}=0.64I _0$

  2. $I _{min}=0.36I _0$

  3. $\dfrac{I _{max}}{I _{min}}=16$

  4. $\dfrac{I _{max}}{I _{min}}=\dfrac{1.64}{0.36}$


Correct Option: A

Path difference between two waves from a coherent sources is 5 nm at a  point P. Wavelength of these waves is 100 $\mathring { A } $. Resultant intensity at point P if intensity of sources is $l _0$ and $4l _0$

  1. Zero

  2. $l _0$

  3. $5l _0$

  4. $3l _0$


Correct Option: B
Explanation:

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

A laser beam can be focussed on an area equal to the square of its wavelength. A He-Ne laser radiates energy at the rate of $1\,nW$ and its wavelength is $632.8\,nm$.The intensity of foucussed beam will be   

  1. $1.5 \times {10^{13}}\,W/{m^2}$

  2. $0.25 \times {10^{4}}\,W/{m^2}$

  3. $3.5 \times {10^{17}}\,W/{m^2}$

  4. None of these


Correct Option: B
Explanation:

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

A He-Ne laser radial co energy 
at the rate$=1nW$
wavelength$=6.32.8nm$
The intensity of focused beam willbe
Area through which energy of beam passes
$=(6.328\times10^{-7})=4\times10^{-13}m^2\I=\cfrac{P}{A}=\cfrac{10^{-9}}{4\times10^{-13}}\ \quad=0.25\times10^4W/m^2$

Ration of maximum and minimum intensities is refrence pattern is 25:1 . The ration of intensities of refring waves is:

  1. 25 : 1

  2. 5 : 1

  3. 6 : 4

  4. 625 : 1


Correct Option: C

Two waves of intensities $I$ and $4I$ superimpose. The minimum and maximum intensities will respectively be

  1. $I,\space 9I$

  2. $3I,\space 5I$

  3. $I,\space 5I$

  4. None of these


Correct Option: A
Explanation:
The Intensity of the wave is directly proportional to the square of its amplitude.
$I \propto A^{2}$;
$I = cA$$^2$;   '$c$' is an arbitrary constant.
So, if a wave with Intensity $I$ has an amplitude of $A(A{ _{1}}$)
A wave with an Intensity of $4I$ would respectively have an amplitude of $2A$($A{ _{2}}$)

If 2 waves with amplitudes $A{ _{1}}$,$A{ _{2}}$ are superimposed, the resultants would be
Maximum of $A{ _{1}}$ + $A{ _{2}} = 3A$ (Constructive Interference)
Minimum of $|A{ _{1}}$ - $A{ _{2}} |   = A $ (Destructive Interference)
The wave of amplitude $3A$ would have an Intensity of $9I$
The wave of amplitude $A$ would have an Intensity of $I$

For a wave displacement amplitude is $10^{-8} m$ density of air $1.3 kg m^{-3}$ velocity in air $340 ms^{-1}$ and frequency is 2000 Hz.The average intensity of wave is

  1. $5.3\times 10^{-4} Wm^{-2}$

  2. $5.3\times 10^{-6} Wm^{-2}$

  3. $3.5\times 10^{-8} Wm^{-2}$

  4. $3.5\times 10^{-6} Wm^{-2}$


Correct Option: A

Two sound waves of equal intensity $I$ superimpose at point $P$ in $90^{\small\circ}$ out of phase. The resultant intensity at point $P$ will be

  1. $4I$

  2. $\sqrt2I$

  3. $2I$

  4. $I$


Correct Option: C
Explanation:

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

Here, $A _1=A _2=A \text{ and } \theta = \pi/2$

So, $A _R=A\sqrt{2(1+\cos(\theta))}=2A\cos(\theta/2)=2A\cos(\pi/4)=\sqrt{2}A$ 

$\Rightarrow I _R=|A _R|^2=2A^2=2I$

A wave of frequency 500$\mathrm { Hz }$ travels between $\mathrm { x }$and $\mathrm { Y }$ and travel a distance of 600$\mathrm { m }$ in 2$\mathrm { sec }$ . between $X$ and $Y .$ How many wavelength are therein distance $X Y$ :

  1. 1000

  2. 300

  3. 180

  4. 2000


Correct Option: A

Four independent waves are represented by the equations :
$y _1 = a _1\  sin\  \omega t$
$y _2 = a _2\ sin\  \omega t$
$y _3 = a _3\ cos\  \omega t$
$y _4 = a _4\ sin\  (\omega t + \pi/3)$ 
Then the waves for which phenomenon of interference will be observed are - 

  1. 1 and 3

  2. 1 and 4

  3. all 1, 2, 3 and 4

  4. None


Correct Option: A

Two sinusoidal plane waves same frequency having intensities $I _0 $ and $ 4I _0 $ are travelling in same direction. The resultant intensity at a point at which waves meet with a phase difference of zero radian is

  1. $ I _0$

  2. $5 I _0$

  3. $9 I _0$

  4. $3 I _0$


Correct Option: C
Explanation:

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

Resultant intensity, $I=I _1+I _2+2\sqrt{I _1I _2} cos\phi $
                                   $= I _0+4I _0+2\sqrt{I _04I _0} cos0^{\circ} \ =9I _0 $

If the ratio of maximum to minimum intensity in beat is 49, then the ratio of amplitudes of two progressive wave trains

  1. 7:1

  2. 4:3

  3. 49:1

  4. 16:9


Correct Option: B
Explanation:

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

$\dfrac{(\sqrt{I _1}+\sqrt{I _2})}{(\sqrt{I _1}-\sqrt{I _2})}=7$

$\sqrt{I _1}+\sqrt{I _2}=7(\sqrt{I _1}-\sqrt{I _2})$

$\dfrac{\sqrt{I _1}}{\sqrt{I _2}}=\dfrac{4}{3}$

$\dfrac{a}{b}=\dfrac{4}{3}$

Here, $a =\sqrt{I _1}$ and $b =\sqrt{I _2}$, where a and b are the amplitudes of the two progressive waves.

If the intensities of two interfering waves be $ I _1 $ and $ I _2  $, the contrast between maximum and minimum intensity is maximum, when

  1. $I _1 > > I _2$

  2. $I _1 < < I _2$

  3. $I _1 = I _2$

  4. either $I _1$ or $I _2$ is zero


Correct Option: C
Explanation:

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

If the phase difference between two sound waves of wavelength $  \lambda  $ is $  60^{\circ} $, the corresponding path difference is

  1. $ \frac{\lambda}{6} $

  2. $ \frac{\lambda}{2} $

  3. $ \lambda 2 $

  4. $ \frac{\lambda}{4} $


Correct Option: A

Equations of stationary and a travelling wave are as follows: $Y _1=sin\, kx\, cos\,\omega t$ and $Y _2=a\, sin\, (\omega t-kx)$. The phase difference between two points $X _1=\dfrac{\pi}{3k}$ and $ X _2=\dfrac{3\pi}{2k}$ are $\phi _1$ and $\phi _2$ respectively for the two waves.The ratio of $\dfrac{\phi _1}{\phi _2}$ is 

  1. 6

  2. 5

  3. 4

  4. 2


Correct Option: A

Two waves of intensities 1 and 4 superimposes. Then the maximum and minimum intensities are :

  1. 9 and 1

  2. 31 and 1

  3. 91 and 31

  4. 61 and 1


Correct Option: A
Explanation:

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


$\dfrac{maximum\ intensity}{minimum\ intensity}=\left (\dfrac{3}{1}  \right )^2=\dfrac{9}{1}$

Two periodic waves of intensities ${I} _{1}$ and ${I} _{2}$ pass through a region at the same time in the same direction. The sum of the maximum and minimum intensities possible is :

  1. ${I} _{1} + {I} _{2}$

  2. ${\left(\sqrt{{I} _{1}} + \sqrt{{I} _{2}}\right)}^{2}$

  3. ${\left(\sqrt{{I} _{1}} - \sqrt{{I} _{2}}\right)}^{2}$

  4. $2\left({I} _{1} + {I} _{2}\right)$


Correct Option: D
Explanation:

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) $

State whether true or false :
The phenomenon of interference is consistent with the law of conservation of momentum.

  1. True

  2. False


Correct Option: B
Explanation:

The correct answer is option(B).

Interference is nothing but simply a phenomena of redistribution of energy. If it is constructive then energy is increased and hence intensity also and if it is destructive then energy is decreased and hence intensity also. So energy is redistributed in constructive and destructive interference but remains conserved.

A travelling wave represented by $y=A\sin { \left( \omega t-kx \right)  } $ is superimposed on another wave represented by $y=A\sin { \left( \omega t+kx \right)  } $. The resultant is 

  1. A standing wave having nodes at$\quad x=\left( n+\cfrac { 1 }{ 2 } \right) \cfrac { \lambda }{ 2 } $, where $n=0,1,2$

  2. A wave travelling along $+x$ direction

  3. /a wavelength travelling along $-x$ direction

  4. a standing wave having nodes at $x=\cfrac { n\lambda }{ 2 } $, where $n=0,1,2$


Correct Option: A
Explanation:

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)

It represents a standing wave.
In the standing wave, there will be nodes (where amplitude is zero) and antinodes  (where amplitude is largest).
From Eq. (i), the positions of nodes are given by
$sin\ kx = 0 \Longrightarrow kx = n\pi; n = 0, 1, 2, ....$

or $\dfrac{2\pi}{\lambda}x = n\pi; 0, 1, 2, ....$

or $x = \dfrac{n\lambda}{2}; n = 0, 1, 2, ...$

In the same way,
From Eq.(i), the positions of antinodes are given by$|sinkx| = 1$
$\Longrightarrow kx = (n + \dfrac{1}{2})\pi ; n = 0, 1, 2, ..... $

or $\dfrac{2\pi x}{\lambda} =  (n + \dfrac{1}{2})\pi ; n = 0, 1, 2, ..... $

or $x =  (n + \dfrac{1}{2})\dfrac{\lambda}{2} ; n = 0, 1, 2, ..... $

The ratio of intensities of two waves that produce interference pattern is 16:1, then the ratio of maximum and minimum intensities in the pattern is :

  1. 25:9

  2. 9:25

  3. 1: 4

  4. 4:1


Correct Option: A
Explanation:

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  \ $  $\dfrac{I _{max}}{I _{min}} = \dfrac{25}{9}$

Consider the superposition of N harmonic waves of equal amplitude and frequency. If N is a very large number determine the resultant intensity in terms of the intensity $\left( { I } _{ 0 } \right)$ of each component wave for the condition when the component waves have identical phases.

  1. ${ NI } _{ 0 }$

  2. ${ N }^{ 2 }{ I } _{ 0 }$

  3. $\sqrt { N } { I } _{ 0 }$

  4. ${ I } _{ 0 }$


Correct Option: A
Explanation:

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

So, the intensities will simply get add to give the resultant intensity
$\Rightarrow (Intensity) _{Resultant}=(I _0+I _0+.....I _0)-N\quad times\ \quad\quad\quad\quad\quad=NI _0$

Two waves having their intensities in the ratio 9:1 produce interference. In the interference pattern, the ratio of maximum to minimum intensity is equal to

  1. 2:1

  2. 9:1

  3. 3:1

  4. 4:1


Correct Option: D
Explanation:

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  \ $  $\dfrac{I _{max}}{I _{min}} = \dfrac{16}{4} = \dfrac{4}{1}$

Assertion - Two sinusoidal waves on the same string exhibit interference.
Reason - these waves, add or cancel out according to the principle of superposition

  1. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

  2. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

  3. Assertion is correct but Reason is incorrect

  4. Both Assertion and Reason are incorrect


Correct Option: A
Explanation:

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.

So, the assertion and reason both are correct and the reason is the correct explanation for the assertion.

Which of the following is true?

  1. Both the light and sound waves exhibit interference

  2. Light waves exhibit interference

  3. Sound waves exhibit interference

  4. Neither sound waves nor light waves exhibit interference


Correct Option: A
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