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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$\sqrt{I _1I _2}$

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, $I=k cos^{2}(\dfrac{2\pi /3}{2})     (\because I=I _{0} cos^{2}(\phi /2))$
         $=k(+1/2)^{2}     (\because cos \pi /3=1/2)$
         $=\dfrac{k}{4}$