Physics

Wave Optics and Diffraction

148 Questions

Wave optics explores light phenomena like interference and diffraction. This hub features problems on Young's double slit experiment, Fresnel biprism, and single slit diffraction patterns. These topics regularly appear in physics sections of competitive exams.

Young's double slitFresnel biprismSingle slit diffractionConstructive interferenceFringe width calculations

Wave Optics and Diffraction Questions

Multiple choice physics dual nature of matter and radiation davisson and germer experiment and its conclusion matter waves wave nature of matter

In Davisson-Germer experiment, intensity was maximum for scattering angle equal to

  1. $40$
  2. $50$
  3. $60$
  4. $70$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

In Davisson -Germer experiment, it was observed that the intensity of the scattered electron beam depends on the scattering angle $\phi$. Also, it Davisson and Germer observed that the maximum intensity was detected when the scattering angle was  $50^o$.

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 phase difference between the light waves reaching third bright fringe from the central fringe will be ($\lambda =6000\mathring {A}$)

  1. $0$
  2. $2\pi$
  3. $4\pi$
  4. $6 \pi$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\because n=3\ \therefore 2n\pi =2\times 3\times \pi \ \quad \quad \quad \quad \quad =6\pi $

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

If two coherent light waves produce minima of fifth order, the path dfference between the waves is

  1. $5 \lambda$
  2. $5 \lambda / { 2 }$
  3. $7 \lambda /{ 2 }$
  4. $9 \lambda / { 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For destructive interference (minima), the path difference must be an odd multiple of lambda/2, specifically (2n-1) * lambda/2. For the fifth order minimum (n=5), the path difference is (2*5 - 1) * lambda/2 = 9*lambda/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 phase difference between the two waves reaching at the location of the third dark fringe is  

  1. $\pi$
  2. $\cfrac{3\pi}{2}$
  3. $5 \pi$
  4. $3 \pi$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The condition for dark fringes in YDSE is path difference = (2n-1) * lambda/2. For the third dark fringe (n=3), path difference = (2*3-1) * lambda/2 = 5*lambda/2. Phase difference phi = (2*pi/lambda) * path difference = (2*pi/lambda) * (5*lambda/2) = 5*pi.

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 constant phase difference between two sources is $\dfrac{\pi }{2}$. The intensity at a point equidistant from the slits in terms of maximum intensity $I _{\circ}$ is :

  1. $I _{\circ}$
  2. $I _{\circ}/2$
  3. $3I _{\circ}/4$
  4. $3I _{\circ}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

If there is no phase difference let
the Intensity equal to maximum Intensity be $I _{0}$
For a phase difference $\theta $ then
Intensity at a point equidistant from the two
slits is given by
$\dfrac{I _{0}}{2}(1+cos^{2}\theta )$
As $\theta =\pi /2$
we get $\dfrac{I _{0}}{2}(1+0)=\dfrac{I _{0}}{2}$
Thus Intensity $=\dfrac{I _{0}}{2}$

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

The maximum intensity produced by two coherent sources of intensity $I _{1}$ and $I _{2}$ constructively will be:

  1. $I _{I}+I _{2}$
  2. $I _{I}^{2}+I _{2}^{2}$
  3. $I _{I}+I _{2}+2\sqrt{I _{1}I _{2}}$

  4. zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If the angle between two coherent sources is $\theta $ then the Intensity after Interference is given by $I _{1}+I _{2}+2\sqrt{I _{1}I _{2}} cos \theta $. For constructive Interference $\theta =2n\pi $ Thus Intensity is $I _{1}+I _{2}+2\sqrt{I _{1}I _{2}}$

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

The intensity ratio for the two interfering beam of light is $\beta$. What is the value of 
$\dfrac{I _{max}-I _{min}}{I _{max}+I _{min}}$ ?

  1. 2$\sqrt{\beta}$
  2. $\dfrac{2\sqrt{\beta}}{1+\beta}$
  3. $\dfrac{2}{1+\beta}$
  4. $\dfrac{1+\beta}{2\sqrt{\beta}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $\cfrac { { I } _{ 1 } }{ { I } _{ 2 } } =\beta $

${ I } _{ max }={ I } _{ 1 }+{ I } _{ 2 }+2\sqrt { { I } _{ 1 }{ I } _{ 2 } } $
${ I } _{ min }={ I } _{ 1 }+{ I } _{ 2 }-2\sqrt { { I } _{ 1 }{ I } _{ 2 } } $
$\therefore \cfrac { { I } _{ max }-{ I } _{ min } }{ { I } _{ max }+{ I } _{ min } } \quad =\cfrac { 4\sqrt { { I } _{ 1 }{ I } _{ 2 } }  }{ 2({ I } _{ 1 }+{ I } _{ 2 }) } \quad =\frac { 2\sqrt { { I } _{ 2 }^{ 2 }\beta  }  }{ { I } _{ 2 }(\beta +1) } \quad =\cfrac { 2\sqrt { \beta  }  }{ \beta +1 } $

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

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

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