Tag: multiple-slit diffraction

Questions Related to multiple-slit diffraction

In Young's double slit experiment, the phase difference between the light waves reaching third bright fringe from the central fringe will be ($\lambda =6000\mathring {A}$)

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

  1. $0$

  2. $2\pi$

  3. $4\pi$

  4. $6 \pi$

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Correct Option: D
Explanation:

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

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 $\frac{I'}{I _0}$ :-

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

  1. 3 : 4

  2. 4 : 3

  3. 2 : 1

  4. 2 : 3

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Correct Option: B

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

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

  1. $5 \lambda$

  2. $5 \lambda / { 2 }$

  3. $7 \lambda /{ 2 }$

  4. $9 \lambda / { 2 }$

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Correct Option: D

In Young's double slit experiment, the phase difference between the two waves reaching at the location of the third dark fringe is  

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

  1. $\pi$

  2. $\cfrac{3\pi}{2}$

  3. $5 \pi$

  4. $3 \pi$

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Correct Option: C

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,

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

  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

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Correct Option: B

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

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

  1. Candle power

  2. Dioptre

  3. Lumen

  4. Intensity of illumination

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Correct Option: D
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.

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

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

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

  2. $ \dfrac{\pi}{ 2}$

  3. $ \pi $

  4. zero

Show answer
Correct Option: C
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.

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 :

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

  1. $I _{\circ}$

  2. $I _{\circ}/2$

  3. $3I _{\circ}/4$

  4. $3I _{\circ}$

Show answer
Correct Option: B
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 $=\dfrac{I _{0}}{2}$

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

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

  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

Show answer
Correct Option: C
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 $I _{1}+I _{2}+2\sqrt{I _{1}I _{2}}$

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

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

  1. 2$\sqrt{\beta}$

  2. $\dfrac{2\sqrt{\beta}}{1+\beta}$

  3. $\dfrac{2}{1+\beta}$

  4. $\dfrac{1+\beta}{2\sqrt{\beta}}$

Show answer
Correct Option: B
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 } $