Tag: oscillations and waves

Quizzes (22) Questions (115)

Questions Related to oscillations and waves

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$\sqrt{I _1I _2}$

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

Young's double slit experiment is conducted with light of wavelength $\lambda $. The intensity of the bright fringe is $I _{o}$ . The intensity at a point, where path difference is $\lambda $ /4 is given by :

  1. $zero$
  2. $I _{o}/8 $
  3. $I _{o}/4 $
  4. $I _{o}/2 $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In YDSE, intensity from both slits are same ,
So, $I _{0}=I _{max}=I+I+2\sqrt{II}=4I$
Now at path difference $\dfrac{\lambda }{4}$,
$\Delta \phi =\dfrac{2\pi }{4}=\dfrac{\pi }{2}$
So, $I=I+I+2\sqrt{\pm I}cos(\dfrac{\pi }{2})$
$=2I+2\sqrt{II}(0)    (\because cos\pi /2=0)$
$=2I$
So, $I=\dfrac{I _{0}}{2}$

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

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference '$\lambda $' is 'K' units. The intensity of light at a point where the path difference is $\dfrac{\lambda}{3} $ is ($\lambda $ being the wavelength of light used)

  1. K/2

  2. K/4

  3. K

  4. K/3
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

phase difference corresponding to path difference of $\lambda /3$ is
$\phi =\dfrac{2\pi }{3}$
So, $I=k cos^{2}(\dfrac{2\pi /3}{2})     (\because I=I _{0} cos^{2}(\phi /2))$
         $=k(+1/2)^{2}     (\because cos \pi /3=1/2)$
         $=\dfrac{k}{4}$

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

In an interference experiment, phase difference for points where the intensity is minimum is (n = 1, 2, 3 ...)

  1. $n \pi$
  2. $(n\, +\, 1) \pi$
  3. $(2n\, +\, 1) \pi$

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

Intensity at a point due to interference of beams of intensities $I _1,I _2$ with a phase difference $\phi$ between them=$I _1+I _2+2\sqrt{I _1I _2}cos\phi$

The value of the resultant intensity is minimum for $cos\phi=-1$
$\implies \phi=(2n+1)\pi$ for positive integer values of $n$