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 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

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
A Correct answer
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. $I _1=I _2$

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

The ratio of intensities of two waves that produce interference pattern is 16:1, then the ratio of maximum and minimum intensities in the pattern is :

  1. 25:9

  2. 9:25

  3. 1: 4

  4. 4:1
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the intensities of the two waves be $I _1$ and $I _2$.
Given :  $I _1:I _2 = 16:1$
Ratio of maximum and minimum intensities  $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{I _1}  +\sqrt{I _2}}{\sqrt{I _1} - \sqrt{I _2}}\bigg)^2$
Or   $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{\frac{I _1}{I _2}}  +1}{\sqrt{\frac{I _1}{I _2}} - 1}\bigg)^2$

Or  $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{16}  +1}{\sqrt{16} - 1}\bigg)^2 = \bigg(\dfrac{4+1}{4-1}\bigg)^2$
$\implies  \ $  $\dfrac{I _{max}}{I _{min}} = \dfrac{25}{9}$

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

Two waves having their intensities in the ratio 9:1 produce interference. In the interference pattern, the ratio of maximum to minimum intensity is equal to

  1. 2:1

  2. 9:1

  3. 3:1

  4. 4:1
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let the intensities of the two waves be $I _1$ and $I _2$.
Given :  $I _1:I _2 = 9:1$
Ratio of maximum and minimum intensities  $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{I _1}  +\sqrt{I _2}}{\sqrt{I _1} - \sqrt{I _2}}\bigg)^2$
Or   $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{\frac{I _1}{I _2}}  +1}{\sqrt{\frac{I _1}{I _2}} - 1}\bigg)^2$

Or  $\dfrac{I _{max}}{I _{min}} = \bigg(\dfrac{\sqrt{9}  +1}{\sqrt{9} - 1}\bigg)^2 = \bigg(\dfrac{3+1}{3-1}\bigg)^2$
$\implies  \ $  $\dfrac{I _{max}}{I _{min}} = \dfrac{16}{4} = \dfrac{4}{1}$

Multiple choice physics wave optics polarization of light polarisation of light polarisation

A point source of monochromatic light is situated at the centre of a circle, what is the phase difference between the light waves passing through the end points of any diameter

  1. $\dfrac{\pi}{2}$
  2. $\pi$
  3. $\dfrac{3\pi}{2}$
  4. $zero$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A point source emits spherical waves. Any points at the same distance from the source, such as the endpoints of a diameter, lie on the same wavefront and thus have zero phase difference.

Multiple choice physics superposition of waves coherence young's double slit experiment interference

In a biprism experiment, the distance of 20 th bright bandfrom the center of the interference pattern is 8$\mathrm { mm }$ . The distance of 30th bright band from the center is

  1. $11.8\mathrm { mm }$
  2. 12$\mathrm { mm }$
  3. 14$\mathrm { mm }$
  4. 16$\mathrm { mm }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given

$\begin{array}{l} 20\beta =8mm \ \beta =\dfrac { 8 }{ { 20 } }  \ Now, \ 30th\, \, \max  ima=30\beta =\dfrac { { 30\times \beta  } }{ { 20 } } =12mm \ 30th\, \min  ima=\dfrac { { \left( { 2\left( { 30 } \right) -1 } \right)  } }{ 2 } \beta  \ =\dfrac { { 59 } }{ 2 } \beta  \ =\dfrac { { 59 } }{ 2 } \times \dfrac { 8 }{ { 20 } }  \ =11.8mm \ Hence,\, option\, A\, is\, the\, correct\, answer. \end{array}$

Multiple choice physics superposition of waves coherence young's double slit experiment interference

Two coherent waves of light will not produce constructive interference if the phase difference between them is

  1. $0^0$
  2. $360^0$
  3. $720^0$
  4. $90^0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Constructive interference for coherent waves takes place when phase difference = $2n\pi$ where n is a integer

So the only case in the options when constructive interference will not take place is $90^\circ$

Answer. D

Multiple choice physics superposition of waves coherence young's double slit experiment interference

Interference pattern can be produced by two identical sources. Here the identical sources mean that

  1. their size is same

  2. their wavelength is same

  3. the intensity of light emitted by them is same

  4. the emplitudes of light waves emitted by them are same
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For observing interference the term identical source means that their wavelength are the same ( i.e., they are coherent).

Multiple choice physics superposition of waves coherence young's double slit experiment interference

Which of the following is not essential for two sources of light in Young's double slit experiment to produce a sustained interference?

  1. Equal wavelength

  2. Equal intensity

  3. Constant phase relationship

  4. Equal frequency
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For sustainable interference wavelength has to be same. So frequency has to be same too. Contact phase relation is also necessary to sustain the interference pattern else the pattern will keep changing so rapidly that we will not see any pattern.

Equal intensity is not a requirement to keep pattern sustained.

Answer. B) equal intensity

Multiple choice physics superposition of waves coherence young's double slit experiment interference

Instead of using two slits, if we use two separate identical sodium lamps in Young's experiment, which of the following will occur?

  1. General illumination

  2. Widely separate interference

  3. Very bright maxima

  4. Very dark minima
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

There will be general illumination as super imposing waves do not have constant phase difference.