Tag: superposition of waves-1: interference and beats

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Questions Related to superposition of waves-1: interference and beats

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

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

  1. $I \propto A$
  2. $I \propto {A^2}$
  3. $I \propto {A^4}$
  4. $I \propto {A^{-1}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

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. $R=3\sqrt{3I}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

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. $I _1=I, I _2=4I$
Reveal answer Fill a bubble to check yourself
B Correct answer
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)

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

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. $4 : 1$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $90^{o}$.
$\theta = 90^{o}$
$I _{2}= kA^{2} \cos^{2} \left( \dfrac{90^{o}}{2} \right)$
$I _{2} = kA^{2} \cos^{2} (45^{o})$
$I _{2}= kA^{2} \left( \dfrac{1}{2} \right)$ ----- $(2)$
$\dfrac{I _{1}}{I _{2}}= \dfrac{kA^{2}}{kA^{2} (1/2)} =\dfrac{2}{1}$
Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

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. $4I _0$
Reveal answer Fill a bubble to check yourself
D Correct answer
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, $\cfrac{I _{new}}{I _0}=\cfrac{4a^2}{a^2}\\implies I _{new}=4I _0$

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

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. $4a^{2}\cos^{2}\delta/2$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

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. $\dfrac{I _{max}}{I _{min}}=\dfrac{1.64}{0.36}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

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. $3l _0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

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
Reveal answer Fill a bubble to check yourself
B Correct answer
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
$=(6.328\times10^{-7})=4\times10^{-13}m^2\I=\cfrac{P}{A}=\cfrac{10^{-9}}{4\times10^{-13}}\ \quad=0.25\times10^4W/m^2$