Tag: physics

Questions Related to physics

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. $4: 1$

Show answer
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

Show answer
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 $

Show answer
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$\sqrt{I _1I _2}$

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