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Interference Of Light - Condition And Types - Practice Questions & MCQ

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Quick Facts

  • Interference of light waves- 1, Interference of light waves- 2 is considered one of the most asked concept.

  • 28 Questions around this concept.

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QuestionMEDIUM

Two waves \mathrm{y}_{1}=\mathrm{A}_{1} \sin \left(\omega \mathrm{t}-\beta_{1}\right) and \mathrm{y}_{2}=\mathrm{A}_{2} \sin \left(\omega \mathrm{t}-\beta_{2}\right) superimpose to form a resultant wave. The amplitude of this resultant wave is given by

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Time of exposure for a photographic print is 10 seconds, when a lamp of 50 cd is placed at 1 m from it. Then another lamp of luminous intensity I is used, and is kept at 2 m from it. If the time of exposure now is 20 s, the value of I (in cd) is

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A parallel beam of white light falls on a thin film whose refractive index is 1.33. If angle of incidence is 52^{\circ}  then thickness of the film for the reflected light to be coloured yellow (\lambda=6000 \AA) most intensively must be

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The interference pattern is obtained with two coherent light sources of intensity ratio n. In the interference pattern, the ratio

\frac{{I_{\max } - I_{\min } }} {{I_{\max } + I_{\min } }}

will be

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In the Young's double-slit experiment, the intensity of light at a point on the screen where the path difference is \lambda is K, (\lambda being the wave length of light used). The intensity at a point where the path difference is \frac{\lambda }{4}, will be:

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In Young's double slit experiment, one of the slits is wider than the other, so that the amplitude of the light from one slit is double of that from the other slit. If Im  be the maximum intensity, the resultant intensity I when they  interfere at phase difference  \phiis given by :

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Two beams of light having intensities I and 4 I interfere to produce a fringe pattern on the screen. Phase difference between the beams is \frac{\pi}{2}  at point a and \pi at point B. Difference of intensities at A and B is

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Intensity of central bright fringe due to interference of two identical coherent monochromatic sources is I. If one of the source is switched off, then intensity of central bright fringe becomes

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In an interference pattern by two identical slits, intensity of central maxima is \mathrm{I}. If one slit is closed, intensity of central maxima changes to \mathrm{I}_{0}. Then \mathrm{I} and \mathrm{I}_{0}  are related by

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Interference fringes from sodium light  (\lambda=5890 \AA) in a double slit experiment have an angular width 0.20^{\circ}, To increase the fringe width by 10 \%, new wavelength should be

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Concepts Covered - 2

Interference of light waves- 1

In order to observe interference in light waves, the following conditions must be met:

  •  The sources must be coherent.
  • The source should be monochromatic (that is, of a single wavelength).

Coherent sources-

Two sources are said to be coherent if they produce waves of the same frequency with a constant phase difference.  

  • The relation between  Phase difference (\Delta \phi )  and Path difference  (\Delta x )

Phase difference (\Delta \phi ):

The difference between the phases of two waves at a point is called phase difference.

\text { i.e. if } y_{1}=a_{1} \sin \omega t \text { and } y_{2}=a_{2} \sin (\omega t+\phi) \text { so phase difference }=\phi

Path difference (\Delta x ): 

The difference in path lengths of two waves meeting at a point is called path difference between the waves at that point.

And The relation between  Phase difference (\Delta \phi )  and Path difference  (\Delta x ) is given as

   \Delta \phi =\frac{2\pi }{\lambda }\Delta x=k\Delta x

where \lambda =wavelength \ of \ waves

 

 

Interference of light waves- 2

Principle of Super Position-

According to the principle of Super Position of waves, when two or more waves meet at a point, then the resultant wave has a displacement (y) which is the algebraic sum of the displacements ( y_1 \ \ and \ \ y_2) of each wave.

i.e y=y_1+y_2

consider two waves with the equations as

 \begin{array}{l}{y_{1}=A_{1} \sin (k x-w t)} \\ {y_{2}=A_{2} \sin (k x-w t+\phi)}\end{array}

where \phi is the phase difference between waves y_1 \ \ and \ \ y_2.

And According to the principle of Super Position of waves

\begin{aligned} y=& y_{1}+y_{2}=A_{1} \sin (k x-w t)+A_{2} \sin (k x-w t+\phi) \\ &=A_{1} \sin (k x-w t)+A_{2}[\sin (k x-w t) \cos \phi+\sin \phi \cos (k x-\omega t)] \\ \Rightarrow y &=\sin (k x-w t)\left[A_{1}+A_{2} \cos \phi\right]+A_{2} \sin \phi \cos (k x-w t) \dots (1) \end{aligned}

Now let

 Acos\theta =A_{1}+A_{2} \cos \phi \\ and \ \ Asin\theta=A_{2} sin \phi

Putting this in equation (1) we get 

y=A \sin (k x-\omega t) \cos \theta+A \sin \theta \cos (k x-\omega t)

thus we get the equation of the resultant wave as

y=A \sin (k x-\omega t+\theta)

where A=Resultant amplitude of two waves

and A= \sqrt{{A_{1}}^{2}+{A_{2}}^{2}+2A_{1}A_{2}\cos \phi}

and \theta=\tan^{-1} (\frac{A_{2} \sin \phi }{A_{1}+A_{2} \cos \phi})

where 

A_{1}= the amplitude of wave 1

A_{2}=  the amplitude of wave 2

  • \begin{array}{c} \ \ A_{\max }={a_{1}+a_{2}} \ \ and \ \ A_{\min }={a_{1}-a_{2}}\end{array}

Resultant Intensity of two waves (I)-

Using I \ \ \alpha \ \ A^2

we get  I= I_{1}+I_{2}+2\sqrt{I_{1}I_{2}}\cos \phi

where 

I_{1}= The intensity of wave 1

I_{2}= The intensity of wave 2

  • \begin{aligned} \ \ I_{\max } &=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \Rightarrow I_{\max } &=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2} \end{aligned}
  • \begin{array}{c}{I_{\min }=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}}} \Rightarrow {I_{\min }=(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}} \end{array}
  • For identical source-

          I_{1}=I_{2}=I_{0} \Rightarrow I=I_{0}+I_{0}+2 \sqrt{I_{0} I_{0}} \cos \phi=4 I_{0} \cos ^{2} \frac{\phi}{2}

  • \text { Average intensity : } I_{a v}=\frac{I_{\max }+I_{\min }}{2}=I_{1}+I_{2}
  • The ratio of maximum and minimum intensities

\frac{I_{\max }}{I_{\min }}=\left(\frac{\sqrt{I_{1}}+\sqrt{I_{2}}}{\sqrt{I_{1}}-\sqrt{I_{2}}}\right)^{2}=\left(\frac{\sqrt{I_{1} / I_{2}}+1}{\sqrt{I_{1} / I_{2}}-1}\right)^2=\left(\frac{a_{1}+a_{2}}{a_{1}-a_{2}}\right)^{2}=\left(\frac{a_{1} / a_{2}+1}{a_{1} / a_{2}-1}\right)^{2}

or

 \sqrt{\frac{I_{1}}{I_{2}}}=\frac{a_{1}}{a_{2}}=\left(\frac{\sqrt{\frac{I_{\max }}{I_{\min }}}+1}{\sqrt{\frac{I_{\max }}{I_{\min }}-1}}\right)

Interference of Light-

It is of the following two types.

1. Constructive interference-

  • When the waves meet a point with the same phase, constructive interference is obtained at that point.

            i.e we will see bright fringe/spot.

  • The phase difference between the waves at the point of observation is  \phi=0^{\circ} \text { or } 2 n \pi

  • Path difference between the waves at the point of observation is \Delta x=n \lambda(i . e . \text { even multiple of } \lambda / 2)

  • The resultant amplitude at the point of observation will be maximum

      \begin{array}{c} i.e \ \ A_{\max }={a_{1}+a_{2}} \\ {\text { If } a_{1}=a_{2}=a_{0} \Rightarrow A_{\max }=2 a_{0}}\end{array}

  • Resultant intensity at the point of observation will be maximum

   \begin{aligned} \ \ i.e \ \ I_{\max } &=I_{1}+I_{2}+2 \sqrt{I_{1} I_{2}} \\ I_{\max } &=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2} \\ \text { If } \quad I_{1} &=I_{2}=I_{0} \Rightarrow I_{\max }=4 I_{0} \end{aligned}

 

2. Destructive interference-

  • When the waves meet a point with the opposite phase, Destructive interference is obtained at that point.

            i.e we will see dark fringe/spot.

  • The phase difference between the waves at the point of observation is 

        \begin{array}{l}{\phi=180^{\circ} \text { or }(2 n-1) \pi ; n=1,2, \ldots} \\ {\text { or }(2 n+1) \pi ; n=0,1,2 \ldots .}\end{array}  

  • Path difference between the waves at the point of observation is \Delta x=(2 n-1) \frac{\lambda}{2}(\text { i.e. odd multiple of } \lambda / 2)

  • The resultant amplitude at the point of observation will be maximum

      \begin{array}{c} i.e \ \ A_{\min }={a_{1}-a_{2}} \\ {\text { If } a_{1}=a_{2} \Rightarrow A_{\min }=0}\end{array}

  • Resultant intensity at the point of observation will be maximum

    \begin{array}{c}{I_{\min }=I_{1}+I_{2}-2 \sqrt{I_{1} I_{2}}} \\ {I_{\min }=(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}} \\ {\text { If } I_{1}=I_{2}=I_{0} \Rightarrow I_{\min }=0}\end{array}

Study it with Videos

Interference of light waves- 1
Interference of light waves- 2

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