Tag: superposition of waves

Quizzes (5) Questions (72)

Questions Related to superposition of waves

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 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 amount of light that falls per unit area held perpendicular to the rays in one second is called

  1. Candle power

  2. Dioptre

  3. Lumen

  4. Intensity of illumination
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Candle power- It is the rating of light output at source. It is the the candle's luminous intensity in a particular direction.

Dioptre- It is the ability of a lens to bend the light rays. A unit of power of lens.
Lumen- The amount of energy emanating from one square meter of surface.
Intensity of illumination - The amount of light falling on per unit area, held perpendicular to the rays in one second.

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 } $