Tag: multiple-slit diffraction

Questions Related to multiple-slit diffraction

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

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

  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}} $

Show answer
Correct Option: C
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 }) }  } $

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?

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

  1. $I _o$

  2. $2I _o$

  3. $3I _o$

  4. $4I _o$

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Correct Option: A
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.

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

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

  1. Phase difference of $3 \pi$

  2. Path difference of $3 \lambda $

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

  4. Path difference of $\dfrac{5\lambda }{2}$

Show answer
Correct Option: D
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} $

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

At two points P and Q on screen in Young's double shit experiment, waves from slits $S _1$ and $S _2$ have a path difference of O and $\frac{\lambda}{4}$ respectively, the ratio of intenstine at P and Q will be: 

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

  1. $3: 2$

  2. $2: 1$

  3. $\sqrt2: 1$

  4. $4: 1$

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Correct Option: A


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)

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

  1. 5 cm

  2. 7 cm

  3. 1 cm

  4. zero

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Correct Option: A
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 $

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 ?

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

  1. 6000 $A^0 $

  2. 5000 $A^0 $

  3. 3000 $A^0 $

  4. 1500 $A^0 $

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Correct Option: A

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

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

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

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

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

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

Show answer
Correct Option: B
Explanation:

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

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

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

  1. $I _{max}$

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

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

  4. zero

Show answer
Correct Option: D
Explanation:

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

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

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

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

  1. $(a _1+a _2)$

  2. $(a _1-a _2)$

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

  4. $(a^2 _1-a^2 _2)$

Show answer
Correct Option: C
Explanation:

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

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

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

  1. I$ _1 + I _2$

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

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

  4. zero

Show answer
Correct Option: C
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$\sqrt{I _1I _2}$