
What is interference? Derive the conditions to obtain constructive and destructive interferences. If a white light source is used in place of a monochromatic source of light in Young’s double slit experiment then what will be the effect on interference fringes?
Answer
548.1k+ views
Hint: To solve the above problem we will define the Young’s double slit experiment and related terms like:
Interference is the phenomenon of non uniform distribution of energy where superimposition of light waves exists.
Young’s experiment demonstrated that light has wavelength nature.
We will also analyse what will happen when white light is used for interference fringe experiment.
Complete step by step solution:
First let’s define all the terms related to interference and Young’s double slit experiment.
Young’s double slit experiment: explained that light has wavelength nature. Young demonstrated the interference of light experimentally.
Interference of light: the phenomenon of non uniform distribution of energy in the medium due to superposition of two light waves is called Interference of light.
Fringe: On the interference of light waves, we obtain alternate dark and bright bands of light called fringes.
Fringe width : the spacing between any two consecutive bright fringes is equal to the width of the width of the dark fringe.
${y_1} = {a_1}\sin \omega t$
${y_2} = {a_2}\sin (\omega t + \phi )$ ($\phi $ is the constant phase difference between the two wave form)
On adding both $y_1$ and $y_2$
$
\Rightarrow y = {y_1} + {y_2} \\
\Rightarrow {a_1}\sin \omega t + {a_2}\sin (\omega t + \phi ) \\
\Rightarrow {a_1}\sin \omega t + {a_2}\sin \omega t\cos \phi + {a_2}\sin \phi \cos \omega t………….\text{(using sin(a + b) identity of trigonometry )}$
$
\Rightarrow ({a_1} + {a_2}\cos \phi )\sin \omega t + {a_2}\sin \phi \cos \omega t \\
\Rightarrow A\cos \theta = ({a_1} + {a_2}\cos \phi )\& A\sin \theta = {a_2}\sin \phi
$................................1
( we have assumed Acos$\theta $ instead of the term in bracket, similarly A sin$\theta $ )
Now we have ,
$
\Rightarrow y = A\cos \theta \sin \omega t + A\sin \theta \cos \omega t \\
\Rightarrow y = A\sin (\theta + \omega t)
$( using sinAcosB + sinB cos A)
The expression we have obtained above is simple harmonic wave with amplitude A and phase difference $\theta $
In equation 1 , we will square and add the both terms.
$
\Rightarrow {(A\cos \theta )^2} = {({a_1} + {a_2}\cos \phi )^2}............2 \\
\Rightarrow {(A\sin \theta )^2} = {({a_2}\sin \phi )^2}.............3
$
Add the two equations
$
\Rightarrow {(A\cos \theta )^2} + {(A\sin \theta )^2} = {({a_1} + {a_2}\cos \phi )^2} + {({a_2}\sin \phi )^2} \\
\Rightarrow {A^2}({\cos ^2}\theta + {\sin ^2}\theta ) = {a_1}^2 + {a_2}^2\cos {\phi ^2} + 2{a_1}{a_2}\cos \phi + {a^2}_2{\sin ^2}\phi $..........(using $(a+b)^2$ = $a^2 + b^2 + 2ab$)
$ \Rightarrow {A^2} = {a_1}^2 + {a_2}^2 + 2{a_1}{a_2}\cos \phi $
(this is the final expression of amplitude)
We also know that square of the amplitude is directly proportional to the intensity of light.
Thus, $I \propto A^2$ ($I$ is the intensity of light)
So, we can write the equation in terms of intensity:
$ \Rightarrow I \propto {a_1}^2 + {a_2}^2 + 2{a_1}{a_2}\cos \phi $....................................4
Condition for constructive interference is, when cos$\phi $ =+1 or $\phi $=0,2$\pi $,4$\pi $
therefore, $\phi $=2n$\pi $, where n=0,1,2.....
It is the condition for constructive interference between the two light waves in terms of phase difference between the waves .
Condition for destructive interference is, when cos$\phi $= -1 or $\phi $= $\pi $, 3$\pi $,5$\pi $,
therefore $\phi $=(2n+1) $\pi $, where n=0,1,2...........
Now the second part of the question:
When monochromatic light is replaced by white light in young’s double slit experiment then we know white light consists of seven different colours having different wavelengths , So each wavelength draws its own pattern of interference.
Note: Fringes can be dark or bright. By dark or bright we mean that area where intensity of light is more forms the bright fringe and the area on the screen where the intensity of the light is less shows dark fringe.
Interference is the phenomenon of non uniform distribution of energy where superimposition of light waves exists.
Young’s experiment demonstrated that light has wavelength nature.
We will also analyse what will happen when white light is used for interference fringe experiment.
Complete step by step solution:
First let’s define all the terms related to interference and Young’s double slit experiment.
Young’s double slit experiment: explained that light has wavelength nature. Young demonstrated the interference of light experimentally.
Interference of light: the phenomenon of non uniform distribution of energy in the medium due to superposition of two light waves is called Interference of light.
Fringe: On the interference of light waves, we obtain alternate dark and bright bands of light called fringes.
Fringe width : the spacing between any two consecutive bright fringes is equal to the width of the width of the dark fringe.
${y_1} = {a_1}\sin \omega t$
${y_2} = {a_2}\sin (\omega t + \phi )$ ($\phi $ is the constant phase difference between the two wave form)
On adding both $y_1$ and $y_2$
$
\Rightarrow y = {y_1} + {y_2} \\
\Rightarrow {a_1}\sin \omega t + {a_2}\sin (\omega t + \phi ) \\
\Rightarrow {a_1}\sin \omega t + {a_2}\sin \omega t\cos \phi + {a_2}\sin \phi \cos \omega t………….\text{(using sin(a + b) identity of trigonometry )}$
$
\Rightarrow ({a_1} + {a_2}\cos \phi )\sin \omega t + {a_2}\sin \phi \cos \omega t \\
\Rightarrow A\cos \theta = ({a_1} + {a_2}\cos \phi )\& A\sin \theta = {a_2}\sin \phi
$................................1
( we have assumed Acos$\theta $ instead of the term in bracket, similarly A sin$\theta $ )
Now we have ,
$
\Rightarrow y = A\cos \theta \sin \omega t + A\sin \theta \cos \omega t \\
\Rightarrow y = A\sin (\theta + \omega t)
$( using sinAcosB + sinB cos A)
The expression we have obtained above is simple harmonic wave with amplitude A and phase difference $\theta $
In equation 1 , we will square and add the both terms.
$
\Rightarrow {(A\cos \theta )^2} = {({a_1} + {a_2}\cos \phi )^2}............2 \\
\Rightarrow {(A\sin \theta )^2} = {({a_2}\sin \phi )^2}.............3
$
Add the two equations
$
\Rightarrow {(A\cos \theta )^2} + {(A\sin \theta )^2} = {({a_1} + {a_2}\cos \phi )^2} + {({a_2}\sin \phi )^2} \\
\Rightarrow {A^2}({\cos ^2}\theta + {\sin ^2}\theta ) = {a_1}^2 + {a_2}^2\cos {\phi ^2} + 2{a_1}{a_2}\cos \phi + {a^2}_2{\sin ^2}\phi $..........(using $(a+b)^2$ = $a^2 + b^2 + 2ab$)
$ \Rightarrow {A^2} = {a_1}^2 + {a_2}^2 + 2{a_1}{a_2}\cos \phi $
(this is the final expression of amplitude)
We also know that square of the amplitude is directly proportional to the intensity of light.
Thus, $I \propto A^2$ ($I$ is the intensity of light)
So, we can write the equation in terms of intensity:
$ \Rightarrow I \propto {a_1}^2 + {a_2}^2 + 2{a_1}{a_2}\cos \phi $....................................4
Condition for constructive interference is, when cos$\phi $ =+1 or $\phi $=0,2$\pi $,4$\pi $
therefore, $\phi $=2n$\pi $, where n=0,1,2.....
It is the condition for constructive interference between the two light waves in terms of phase difference between the waves .
Condition for destructive interference is, when cos$\phi $= -1 or $\phi $= $\pi $, 3$\pi $,5$\pi $,
therefore $\phi $=(2n+1) $\pi $, where n=0,1,2...........
Now the second part of the question:
When monochromatic light is replaced by white light in young’s double slit experiment then we know white light consists of seven different colours having different wavelengths , So each wavelength draws its own pattern of interference.
Note: Fringes can be dark or bright. By dark or bright we mean that area where intensity of light is more forms the bright fringe and the area on the screen where the intensity of the light is less shows dark fringe.
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