Tag: superposition of waves

Quizzes (5) Questions (72)

Questions Related to superposition of waves

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

Let ${a _1}$ and ${a _2}$ be the amplitudes of two light waves of same frequency and ${\alpha _1}$ and ${\alpha _2}$ be their initial phases. The resultant amplitude due to the superposition of two light waves is

  1. $R = \sqrt {a _1^2 + a _2^2 + 2{a _1}{a _2}} $

  2. $R = {a _1} - {a _2}$

  3. $R = \sqrt {a _1^2 + a _2^2 + 2{a _1}{a _2}\cos \left( {{\alpha _1} - {\alpha _2}} \right)} $

  4. $R = \sqrt {a _1^2 + a _2^2 - 2{a _1}{a _2}} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The angle between two light waves $={ \alpha  } _{ 1 }-{ \alpha  } _{ 2 }$

Resultant $=\sqrt { { a } _{ 1 }^{ 2 }+{ a } _{ 2 }^{ 2 }+2{ a } _{ 1 }.{ a } _{ 2 }\cos { ({ \alpha  } _{ 1 }-{ \alpha  } _{ 2 }) }  } $

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

In the interference of waves from two sources of intensities $I _o$ and $4I _o$, the intensity at a point where the phase difference is $\pi$, is?

  1. $I _o$

  2. $2I _o$

  3. $3I _o$

  4. $4I _o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$I=I _1+I _2+2\sqrt{I _1I _2}\cos\theta =I _o+4I _o+2\sqrt{(I _o\times 4I _o)}\cos\pi =I _o$
Hence (A) is correct.

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

In Young's double slit experiment, when two light waves form third minimum, they have 

  1. Phase difference of $3 \pi$

  2. Path difference of $3 \lambda $

  3. Phase difference of $\dfrac{5\pi}{2}$

  4. Path difference of $\dfrac{5\lambda }{2}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For minima , path difference=$\dfrac{(2n-1)\lambda}{2}$........(1)

          $n=3$
 so, putting in (1),
we get, path difference=$\dfrac{5\lambda}{2}$

Phase diffrence $\dfrac{\Delta\phi}{2 \pi}=\dfrac {\Delta X}{\lambda}$
 
$\Delta \phi =2\pi \dfrac {\Delta X}{\lambda} $

 $\Delta \phi =2\pi \dfrac {5 \lambda /2}{\lambda} =5 \pi$

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics


The displacement of two interfering light wave are $ y _1 = 4 sin \omega t and y _2 = 3 cos(\omega t) $
The amplitude of the resultant wave is and $ y _2 $ are:(in CGS system)

  1. 5 cm

  2. 7 cm

  3. 1 cm

  4. zero
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given that,

$ {{y} _{1}}=4\sin \omega t $

$ {{y} _{2}}=3\cos \omega t $

Amplitude of first and second wave is 4 cm and 3 cm. So, the amplitude of resultant wave is

$ {{y}^{'}}=\sqrt{{{(4)}^{2}}+{{(3)}^{2}}} $

$ =5\,cm $

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

A light wave is incident normally over slit of width $24\times 10^{-5}$ cm. The angular position of second dark fringe from the central maximum is 30$^{0}$. What is the wavelength of light ?

  1. 6000 $A^0 $

  2. 5000 $A^0 $

  3. 3000 $A^0 $

  4. 1500 $A^0 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For single slit diffraction, the condition for dark fringes is a*sin(theta) = n*lambda. Given a = 24*10^-5 cm = 2400 nm, n=2, theta=30 degrees. 2400 * sin(30) = 2 * lambda => 2400 * 0.5 = 2 * lambda => 1200 = 2 * lambda => lambda = 600 nm = 6000 Angstrom.

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

Two coherent sources of intensity ratio of interfere in interference parteren $\frac { \mathrm { I } _ { \max } - \mathrm { I } _ { \min } } { \mathrm { I } _ { \max } + \mathrm { I } _ { \min } }$ is equal to

  1. $\frac { 2 \alpha } { 1 + \alpha }$

  2. $\frac { 2 \sqrt { a } } { 1 + \alpha }$

  3. $\frac { 2 \alpha } { 1 \sqrt { \alpha } }$

  4. $\frac { 1 + \alpha } { 2 \alpha }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} { { { I } } _{ \max   } }={ \left( { \sqrt { { I _{ 1 } } } +\sqrt { { I _{ 2 } } }  } \right) ^{ 2 } } \ ={ I _{ 1 } }+{ I _{ 2 } }+2\sqrt { { I _{ 1 } }{ I _{ 2 } } }  \ { I _{ \min   } }={ I _{ 1 } }+{ I _{ 2 } }-2\sqrt { { I _{ 1 } }{ I _{ 2 } } }  \ \therefore \dfrac { { { { { I } } _{ \max   } }-{ { { I } } _{ \min   } } } }{ { { I _{ \max   } }+{ { { I } } _{ \min   } } } } =\dfrac { { 4\sqrt { { I _{ 1 } }{ I _{ 2 } } }  } }{ { 2\left( { { I _{ 1 } }+{ I _{ 2 } } } \right)  } }  \ =\dfrac { { 2\sqrt { { I _{ 1 } }{ I _{ 2 } } }  } }{ { \left( { { I _{ 1 } }+{ I _{ 2 } } } \right)  } }  \ { I _{ 1 } }=1 \ { I _{ 2 } }=\alpha  \ =\dfrac { { 2\sqrt { \alpha  }  } }{ { 1+\alpha  } }  \ \therefore \, \, Option\, \, \left( B \right) \, \, is\, \, correct\, . \end{array}$

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

In Young's double slit experiment if the maximum intensity of light is $I _{max}$, then the intensity at path difference $\dfrac{\lambda}{2}$ will be

  1. $I _{max}$

  2. $\displaystyle\frac{I _{max}}{2}$

  3. $\displaystyle\frac{I _{max}}{4}$

  4. zero
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Destructive interference occurs when the difference is an odd multiple of $\pi$ ,

for path difference of $\lambda/2 $ , $\phi =\pi$

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

Two coherent waves are represented by $y _1=a _1\cos\omega t$ and $y _2=a _2\cos\omega t$. The maximum intensity due to interference will be proportional to

  1. $(a _1+a _2)$

  2. $(a _1-a _2)$

  3. $(a^2 _1+a^2 _2)$

  4. $(a^2 _1-a^2 _2)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

 $intensity \ \alpha \ (amplitude)^{2}$
so maximum intensity is proportional to $a^{2} _{1}+a^{2} _{2}$
option $C$ is correct 

Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

The max. intensity produced by two coherent sources of intensity  $I _2$ and $I _2$ will be 

  1. I$ _1 + I _2$

  2. $ I _1^2 + I _2^2$

  3. $ I _1 + I _2$ + 2$\sqrt{I _1I _2}$

  4. zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As R$^{2}$ = a$^{2}$ + b$^{2}$ + 2 ab cos $\phi$
$\therefore$ I$ _{max}$ = I$ _1$ + I$ _2$ + 2$\sqrt{I _1I _2}$ cos 0$^{o}$ 
= I$ _1$ + I$ _2$ + 2$\sqrt{I _1I _2}$