Tag: superposition of waves

Questions Related to superposition of waves

Two sources of light are said to be coherent, when they give light waves of same

  1. amplitude and phase

  2. wavelength and constant phase difference

  3. intensity and wavelength

  4. phase and speed


Correct Option: B
Explanation:

In physics, two wave sources are perfectly coherent if they have a constant phase difference and the same frequency (amplitude may be different).

As $c $ be the speed of light which is constant.
Using, $c= \nu \lambda$
Now same $\nu$ gives same $\lambda.$ for the two light sources.
Example: $y _1= A _1 sin wt$  and $y _2=A _2 sin (wt+\phi) $ where $\phi$ is constant.

For interference to take place

  1. sources must be coherent

  2. sources must have same amplitude

  3. waves should travel in opposite directions

  4. sources must have same frequency


Correct Option: A,D
Explanation:

For interference to take place source must be coherent, which implies:

  • Same frequency for the two waves
  • Constant phase difference.
Options A and D match.

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

  1. Nearly the same frequency

  2. The same frequency

  3. Different wavelengths

  4. The same frequency and having a definite phase relationship


Correct Option: D
Explanation:
As for an Interference, Two Sources must be Coherent, hence the two sources must have the same frequency and a definite phase relationship.

A light of wavelength $400\overset{o}{A}$ after travelling a distance of $2\mu m$ produces a phase change of:

  1. Zero

  2. $3\pi$

  3. $\displaystyle\frac{\pi}{2}$

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


Correct Option: A
Explanation:

The number of wavelengths the wave travels is $=\dfrac{l}{\lambda}$ $=\dfrac{2\times 10^{-6}}{400\times 10^{-10}}$ $=50$     which is an integer.

Hence, the point is at a distance which is integral multiple of wavelength.
Thus the phase difference between the points is zero.

To demonstrate the phenomenon of interference of sound, we need:

  1. two sources, which emit sound of exactly the same frequency

  2. two sources, which emit sound of exactly the same frequency and have a definite phase relationship

  3. two sources, which emit sound of exactly the same frequency and have a varying phase relationship

  4. two sources, which emit sound of exactly the same wavelength


Correct Option: B
Explanation:

For demonstration of interference of sound we need two coherent  source, of same frequency with constant phase difference

$P _1=P _{01}  sin  (Kx-\omega t)$

$P _2=P _{02}  sin  (k(x+\Delta x)-\omega t)$

$=P _{02}  sin  (Kx-\omega t+\delta )$

$\delta =K\Delta x=\dfrac{2\pi \Delta x}{\lambda }$

Two coherent sources of intensity ratio $\beta$ interfere. Then the value of $\displaystyle \left( {\frac{{{I _{\max }} - {I _{\min }}}}{{{I _{\max }} + {I _{\min }}}}} \right)$ is:

  1. $\dfrac{{1 + \beta }}{{\sqrt \beta }}$

  2. $\sqrt {\left( {\dfrac{{1 + \beta }}{\beta }} \right)} $

  3. $\dfrac{{1 + \beta }}{{2\sqrt \beta }}$

  4. $\dfrac{{2\sqrt \beta }}{{1 + \beta }}$


Correct Option: D
Explanation:

$\displaystyle \frac{I _2}{I _1}=\beta$                        $\displaystyle \frac{I _{min}}{I _{max}}=\frac{\beta - 2 \sqrt{\beta}+1}{\beta + 2 \sqrt{\beta +1}}$
$\frac{A _2}{A _1} =\sqrt{\beta}$
$\displaystyle \frac{A _2 -A _1}{A _2 + A _1}=\frac{\sqrt{\beta}-1}{\sqrt{\beta}+1}$     $\displaystyle \frac{I _{max}- I _{min}}{I _{max}+ I _{min}}=\frac{(2)}{(2)} \left [ \frac{2 \sqrt{\beta}}{\beta +1}\right ]$

Coherent sources for studies in interference of light are obtained from.

  1. Two sources derived from a single source of light having a constant phase difference

  2. Two independent sources of light having a varying phase difference

  3. Two independent sources of light having a constant phase difference

  4. None of the above


Correct Option: A

Interference fringes were produced in Young's double slit experiment using light of wavelength $5000\overset{o}{A}$. When a film of thickness $2.5\times 10^{-3}$cm was placed in front of one of the slits, the fringe pattern shifted by a distance equal to $20$ fringe-widths. The refractive index of the material of the film is?

  1. $1.25$

  2. $1.35$

  3. $1.4$

  4. $1.5$


Correct Option: C
Explanation:

$n\lambda =\left( \mu -1 \right) t$\ $20\times 5\times  { 10 }^{ -7 }=\left( \mu -1 \right) 25\times { 10 }^{ -6 }$\ $\mu =1.4$

True & False Statement Type
$S _1$ : In an elastic collision initial and final K.E. of system will be same.
$S _2$ : In a pure L-C Circuit average energy stored in capacitor is zero.
$S _3$ : In YDSE coherent sources are formed by division of wave front method.
$S _4$ : If a physical Quantity is quantized then it must be integral multiple of its lowest value.

  1. FFTF

  2. TTFT

  3. FTFT

  4. TFTT


Correct Option: D
Explanation:

S1: In an elastic collision the kinetic energy is conserved so the initial and final K.E. of system will be same. (T)
S2: In a pure L-C Circuit average energy stored in capacitor always oscillate between a maximum and a minimum. So not zero always.(F)
S3:In YDSE coherent sources are formed by division of wavefront method from a single source of light. (T)
S4: If a physical Quantity is quantized then it will be the integral multiple of its lowest value (T)

In Young's experiment fourth bright fringe produced by light of $5000\overset{o}{A}$ superposes on the fifth bright fringe of an unknown wavelength. The unknown wavelength is _________$\overset{o}{A}$.

  1. $4000$

  2. $6000$

  3. $5000$

  4. $8000$


Correct Option: A
Explanation:
$ \beta = \dfrac{ \lambda \times D}{d} $
$4\beta _1=5\beta _2$
$4\times 5000\dfrac{D}{d}=5\times \lambda\dfrac{D}{d}$
$\lambda =4000\overset{o}{A}$.