Question

(a) Consider two coherent sources ${S_1}$ and ${S_2}$producing monochromatic waves to produce interference patterns. Let the displacement of the wave produced by ${S_1}$ be given by ${Y_1} = a\cos \omega t$ and the displacement by ${S_2}$ is given by ${Y_2} = a\cos \left( {\omega t + \phi } \right)$ . Find out the expression for the amplitude of the resultant displacement at a point and show that the intensity at that point will be $I = 4{a^2}{\cos ^2}\dfrac{{\Delta \phi }}{2}$. Hence establish the conditions for constructive and destructive interference.
(b) What is the effect on the interference fringes in Young's double slit experiment when
(i) The width of the source slit is increased;
(ii) The monochromatic source is replaced by a source of white light

Answer 1

The resultant displacement is the sum of the displacements of the individual waves of the two sources. When the width of slits is increased in YDSE, more light passes through. Think what property of light is altered now and how it will affect the patterns

Complete step by step solution:
The resultant displacement at a point can be expressed as the sum of the displacements of the individual waves of the two sources.
Therefore,
${Y_{res}} = {Y_1} + {Y_2}$
Now we put the respective values given in the question
${Y_{res}} = a\cos \omega t + a\cos \left( {\omega t + \phi } \right)$
Taking common terms together we get
${Y_{res}} = a\left( {\cos \omega t + \cos \left( {\omega t + \phi } \right)} \right)$
Now we use the formula $cosx + \operatorname{cosy} = 2\cos \left( {\dfrac{{x + y}}{2}} \right)\cos \left( {\dfrac{{x - y}}{2}} \right)$
${Y_{res}} = 2a\cos \dfrac{{2\omega t + \phi }}{2}\cos \dfrac{\phi }{2}$
The resultant displacement is of the form $Y = a'\cos \left( {\omega 't + \varphi '} \right)$
Comparing the above two equations we get that the resultant amplitude will be equal to
${A_{res}} = 2a\cos \dfrac{\phi }{2}$
This is the required resultant displacement.
We know that intensity of the resultant light at the point is w=equal to the square of amplitude, Therefore.
$I = {\left( {{A_{res}}} \right)^2}$
$\Rightarrow I = {\left( {2a\cos \dfrac{\phi }{2}} \right)^2}$
$\Rightarrow I = 4{a^2}{\cos ^2}\dfrac{\phi }{2}$
Hence proved.
For Constructive interference the intensity of light should be maximum
This means that
${\cos ^2}\dfrac{{\Delta \phi }}{2} = 1$
Or by simple trigonometric
$\dfrac{{\Delta \phi }}{2} = n\pi$ Where $n$ is any integer
$\Delta \phi = 2n\pi$
For Destructive interference the intensity of light should be minimum
This means that
${\cos ^2}\dfrac{{\Delta \phi }}{2} = 0$
Or by simple trigonometric
$\dfrac{{\Delta \phi }}{2} = \left( {2n + 1} \right)\dfrac{\pi }{2}$ Where $n$ is any integer
$\Delta \phi = \left( {2n + 1} \right)\pi$
The required conditions for:
Constructive Interference is $\Delta \phi = 2n\pi$ or even multiples of $\pi$
Destructive Interference is $\Delta \phi = \left( {2n + 1} \right)\pi$ or odd multiples of $\pi$

(b) (i) When the width of the slit is increased, intensity of the incident light waves increases and hence, the intensity of the fringes formed increases. However, the fringes formed are not very distinctive and increasing the slit width too much spreads the fringes in a greater area hence they become distorted.
(ii) When a monochromatic light source is replaced by a white light source, fringes formed at a point consist of only a single color (wavelength). Multiple fringes of various colors are formed. Central fringe remains white as all colors constructively interfere there.

Note: The fridge properties in YDSE depend on the wavelength of light. Since white light consists of different colors each having different wavelengths, we will see separate fringe patterns for all the colors. The intensity of the fringes is uniform in case of YDSE but this is not true in the case of single slit.