Tag: superposition of waves

Questions Related to superposition of waves

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 :

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

  1. $zero$

  2. $I _{o}/8 $

  3. $I _{o}/4 $

  4. $I _{o}/2 $

Show answer
Correct Option: D
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, $I=\dfrac{I _{0}}{2}$

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)

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

  1. K/2

  2. K/4

  3. K

  4. K/3

Show answer
Correct Option: B
Explanation:

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

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

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

  1. $n \pi$

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

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

  4. zero

Show answer
Correct Option: C
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 $n$

Two identical light waves, propagating in the same direction, have a phase difference $\delta $. After they superpose the intensity of the resulting wave will be proportional to

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

  1. $\cos { \delta } $

  2. $\cos { \left( \dfrac { \delta }{ 2 } \right) } $

  3. $\cos ^{ 2 }{ \left( \dfrac { \delta }{ 2 } \right) } $

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

Show answer
Correct Option: C
Explanation:

Maximum intensity,
     $I=4{ I } _{ 0 }\cos ^{ 2 }{ \left( \dfrac { \delta  }{ 2 }  \right)  } $
$\Rightarrow I\propto \cos ^{ 2 }{ \left( \dfrac { \delta  }{ 2 }  \right)  } $

In the Young's double slit experiment, the resultant intensity at a point on the screen is 75% of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is 

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

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

  2. $\frac {\pi}{4}$

  3. $\frac {\pi}{3}$

  4. $\frac {\pi}{2}$

Show answer
Correct Option: C
Explanation:

The correct answer is option(C).

We know, The resultant intensity,
$I _R=4I _{max}cos^2\left( \frac \phi 2\right)$
$or, cos^2\left( \frac \phi 2\right)=\frac {I _R}{4I _max}=\frac {0.75}4=0.1875$
$\Rightarrow cos\left(\frac \phi 2\right)=\sqrt {0.1875}=0.5$
$\Rightarrow \frac \phi 2 = cos _{-1}(0.5)=\frac \pi 3$

Two beams of light having intensities I and 4 I interface to produce a fringe pattern on a screen. The phase difference between the beams is $\dfrac { \pi  }{ 2 }$ at point A and $\pi$ at point B, Then the difference between the resultant intensities at A and B is

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

  1. 2 I

  2. 4 I

  3. 5 I

  4. 7I

Show answer
Correct Option: B
Explanation:

Resultant intensity at a point, $I _R= I _1+ I _2 +2\sqrt{I _1I _2} cos \phi$

where $I _1$ and $I _2$ be the intensities of two sources and $\phi$ be the phase differences of the sources at that point.
Here, $I _1= I$ and $I _2=4I$

At point $A$, $\phi= \dfrac{\pi}2$
So,Resulant intensity at point $A$,   $I _A= 5I+ 2\sqrt{4I^2}cos \dfrac{\pi}2= 5I$   

At point $B$, $\phi=\pi$

So,Resulant intensity at point $B$,   $I _B= 5I+ 2\sqrt{4I^2}cos\pi= 5I-4I= I$

Hence, Required difference between the intensities at $A$ and $B= I _A-I _B= 5I-I= 4I$

In Young's double slit experiment, the two slits act as coherent sources of waves of equal amplitude $A$ and wavelength $\lambda$. In another experiment with the same arrangement, the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the middle point of the screen in the first case is $I _1$ and in the second case is $I _2$, then the ratio $I _1/I _2$ is: 

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

  1. $0.5$

  2. $4$

  3. $2$

  4. $1$

Show answer
Correct Option: A

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen.If the phase difference between the beams is $\dfrac{\pi }{2}$ at point A and $\pi$ at point B then the difference between the resultant intensities at A and B is 

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

  1. 4I

  2. 2I

  3. 5I

  4. 7I

Show answer
Correct Option: A
Explanation:

The resultant intensity is given by:
$I={I} _{1}+{I} _{2}+2\sqrt{{I} _{1}{I} _{2}}\cos{\phi}$
Thus difference is given by:
${I} _{A}-{I} _{B}=2\sqrt{{I} _{1}{I} _{2}}(\cos{\dfrac{\pi}{2}}-\cos{\pi})=2\sqrt{4{I}^{2}}\times1=4I$

To demonstrate the phenomenon of interference we require two sources which emit radiation of

physics superposition of waves coherence young's double slit experiment interference

  1. nearly the same frequency

  2. the same frequency

  3. different wavelength

  4. the same frequency and having a definite phase relationship

Show answer
Correct Option: D

Which of the following is not an essential condition for interference?

physics superposition of waves coherence young's double slit experiment interference

  1. The two interfering waves must propagate in almost the same direction

  2. The waves must have the same period and wavelength

  3. The amplitudes of the two waves must be equal

  4. The two interfering beams of light must originate from the same source

Show answer
Correct Option: A
Explanation:

When two waves are propagate in same direction