Physics

Wave Optics and Diffraction

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

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

In an interference experiment, phase difference for points where the intensity is minimum is (n = 1, 2, 3 ...)

  1. $n \pi$
  2. $(n\, +\, 1) \pi$
  3. $(2n\, +\, 1) \pi$

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

Intensity at a point due to interference of beams of intensities $I _1,I _2$ with a phase difference $\phi$ between them=$I _1+I _2+2\sqrt{I _1I _2}cos\phi$

The value of the resultant intensity is minimum for $cos\phi=-1$
$\implies \phi=(2n+1)\pi$ for positive integer values of $n$

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

Two identical light waves, propagating in the same direction, have a phase difference $\delta $. After they superpose the intensity of the resulting wave will be proportional to

  1. $\cos { \delta } $
  2. $\cos { \left( \dfrac { \delta }{ 2 } \right) } $
  3. $\cos ^{ 2 }{ \left( \dfrac { \delta }{ 2 } \right) } $
  4. $\cos ^{ 2 }{ \delta } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Maximum intensity,
     $I=4{ I } _{ 0 }\cos ^{ 2 }{ \left( \dfrac { \delta  }{ 2 }  \right)  } $
$\Rightarrow I\propto \cos ^{ 2 }{ \left( \dfrac { \delta  }{ 2 }  \right)  } $

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

In the Young's double slit experiment, the resultant intensity at a point on the screen is 75% of the maximum intensity of the bright fringe. Then the phase difference between the two interfering rays at that point is 

  1. $\frac {\pi}{6}$
  2. $\frac {\pi}{4}$
  3. $\frac {\pi}{3}$
  4. $\frac {\pi}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The correct answer is option(C).

We know, The resultant intensity,
$I _R=4I _{max}cos^2\left( \frac \phi 2\right)$
$or, cos^2\left( \frac \phi 2\right)=\frac {I _R}{4I _max}=\frac {0.75}4=0.1875$
$\Rightarrow cos\left(\frac \phi 2\right)=\sqrt {0.1875}=0.5$
$\Rightarrow \frac \phi 2 = cos _{-1}(0.5)=\frac \pi 3$

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

Two beams of light having intensities I and 4 I interface to produce a fringe pattern on a screen. The phase difference between the beams is $\dfrac { \pi  }{ 2 }$ at point A and $\pi$ at point B, Then the difference between the resultant intensities at A and B is

  1. 2 I

  2. 4 I

  3. 5 I

  4. 7I
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Resultant intensity at a point, $I _R= I _1+ I _2 +2\sqrt{I _1I _2} cos \phi$

where $I _1$ and $I _2$ be the intensities of two sources and $\phi$ be the phase differences of the sources at that point.
Here, $I _1= I$ and $I _2=4I$

At point $A$, $\phi= \dfrac{\pi}2$
So,Resulant intensity at point $A$,   $I _A= 5I+ 2\sqrt{4I^2}cos \dfrac{\pi}2= 5I$   

At point $B$, $\phi=\pi$

So,Resulant intensity at point $B$,   $I _B= 5I+ 2\sqrt{4I^2}cos\pi= 5I-4I= I$

Hence, Required difference between the intensities at $A$ and $B= I _A-I _B= 5I-I= 4I$

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 two slits act as coherent sources of waves of equal amplitude $A$ and wavelength $\lambda$. In another experiment with the same arrangement, the two slits are made to act as incoherent sources of waves of same amplitude and wavelength. If the intensity at the middle point of the screen in the first case is $I _1$ and in the second case is $I _2$, then the ratio $I _1/I _2$ is: 

  1. $0.5$
  2. $4$
  3. $2$
  4. $1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For coherent sources, intensity at the center with equal amplitudes A is I1 = 4A^2. For incoherent sources, intensities simply add up, so I2 = A^2 + A^2 = 2A^2. The ratio I1 / I2 is therefore 4A^2 / 2A^2 = 2.

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

Two beams of light having intensities $I$ and $4I$ interfere to produce a fringe pattern on a screen.If the phase difference between the beams is $\dfrac{\pi }{2}$ at point A and $\pi$ at point B then the difference between the resultant intensities at A and B is 

  1. 4I

  2. 2I

  3. 5I

  4. 7I
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The resultant intensity is given by:
$I={I} _{1}+{I} _{2}+2\sqrt{{I} _{1}{I} _{2}}\cos{\phi}$
Thus difference is given by:
${I} _{A}-{I} _{B}=2\sqrt{{I} _{1}{I} _{2}}(\cos{\dfrac{\pi}{2}}-\cos{\pi})=2\sqrt{4{I}^{2}}\times1=4I$

Multiple choice evs light and shadow dual nature of light nature and sources of light theories on light

Light of wavelength $\lambda $ from a point source falls on a small circular obstacle of diameter d. Dark and bright circular rings around a central bright spot are formed on a screen beyond the obstacle. The distance between the screen and obstacle is D. Then , the condition for the formation of rings, is

  1. $\sqrt { \lambda } \approx \cfrac { { d } }{ 4D } $
  2. $\lambda \approx \cfrac { { d }^{ 2 } }{ 4D } $
  3. $d\approx \cfrac { { \lambda }^{ 2 } }{ D } $
  4. $\lambda \approx \cfrac { D }{ 4 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The condition for diffraction rings around a circular obstacle (Poisson spot) is related to the Fresnel zone plate condition, where lambda is approximately d^2 / 4D.

Multiple choice physics superposition of waves-1: interference and beats interference of sound waves superposition and interference of sound waves properties of sound waves

If the intensities of two interfering waves be $ I _1 $ and $ I _2  $, the contrast between maximum and minimum intensity is maximum, when

  1. $I _1 > > I _2$
  2. $I _1 < < I _2$
  3. $I _1 = I _2$
  4. either $I _1$ or $I _2$ is zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$I _{max}=(\sqrt{I _1}+\sqrt{I _2})^2$
$I _{min}=(\sqrt{I _1}-\sqrt{I _2})^2$
Contrast is maximum, when $I _{min}=0$ ie. $I _1=I _2$