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 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

In Y.D.S.E. two waves of equal intensity produces an intensity $I _0$ at the centre but at a point where path difference is $\frac{\lambda}{6}$ intensity is I'. Then find the ratio $\frac{I'}{I _0}$ :-

  1. 3 : 4

  2. 4 : 3

  3. 2 : 1

  4. 2 : 3
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The phase difference delta phi = (2 pi / lambda) * path difference. For path difference lambda/6, delta phi = pi/3 radians (60 degrees). Intensity I = I_max * cos^2(delta phi / 2), which gives I' = I_0 * cos^2(pi/6) = I_0 * (3/4), so the ratio I'/I_0 is 3:4.

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 the above question, the intensity of the waves reaching a point P far away on the x-axis from each of the four sources is almost the same and equal to $I _0.$ Then,

  1. If $d=\lambda /4,$ the intensity at P is $4I _0.$
  2. If $d=\lambda /6,$ the intensity at P is $3I _0.$
  3. If $d=\lambda /2,$ the intensity at P is $3I _0.$

  4. None of these is true
Reveal answer Fill a bubble to check yourself
B Correct answer
Multiple choice multiple-slit diffraction the principle of superposition of waves superposition of waves oscillations and waves physics

The phase difference between two waves from successive half period zones or strips is :

  1. $\dfrac{\pi}{ 4}$
  2. $ \dfrac{\pi}{ 2}$
  3. $ \pi $

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

The half period zone is provided by an optical device known as zone plate. It is simply a plane parallel gloss plate having concentric circles of radii accurately proportional to the square roots of the consecutive natural numbers 1,2,3 ... etc. The area is given by $\pi \gamma ^{2}$. Hence the
areas are $\pi , 2\pi , 3\pi , 4\pi$,---
The phase difference is thus $\pi$ between each successive half period zones of strips.

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

Let ${a _1}$ and ${a _2}$ be the amplitudes of two light waves of same frequency and ${\alpha _1}$ and ${\alpha _2}$ be their initial phases. The resultant amplitude due to the superposition of two light waves is

  1. $R = \sqrt {a _1^2 + a _2^2 + 2{a _1}{a _2}} $
  2. $R = {a _1} - {a _2}$
  3. $R = \sqrt {a _1^2 + a _2^2 + 2{a _1}{a _2}\cos \left( {{\alpha _1} - {\alpha _2}} \right)} $
  4. $R = \sqrt {a _1^2 + a _2^2 - 2{a _1}{a _2}} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The angle between two light waves $={ \alpha  } _{ 1 }-{ \alpha  } _{ 2 }$

Resultant $=\sqrt { { a } _{ 1 }^{ 2 }+{ a } _{ 2 }^{ 2 }+2{ a } _{ 1 }.{ a } _{ 2 }\cos { ({ \alpha  } _{ 1 }-{ \alpha  } _{ 2 }) }  } $

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

In the interference of waves from two sources of intensities $I _o$ and $4I _o$, the intensity at a point where the phase difference is $\pi$, is?

  1. $I _o$
  2. $2I _o$
  3. $3I _o$
  4. $4I _o$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$I=I _1+I _2+2\sqrt{I _1I _2}\cos\theta =I _o+4I _o+2\sqrt{(I _o\times 4I _o)}\cos\pi =I _o$
Hence (A) is correct.

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 $