Question

(a) What are coherent sources of light? State two conditions for two light sources to be coherent.
(b) Derive a mathematical expression for the width of the interference fringes obtained in Young’s double slit experiment with the help of a suitable diagram.
(c) If $s$ is the size of the source and $b$ its distance from the plane of the two slits, what should be the criterion for the interference fringes to be seen?

Answer 1

Hint:Try to visualize the diagram and use different approximations to solve the question. General geometry can be used. The observations of Young’s double-slit experiment will give a perspective about the occurrence of interference. Coherency is the key to implement the experiment successfully.

Complete step by step answer:
(a) When two light sources produce waves having the same frequency and have a constant phase difference with time, they are called to be coherent. The two conditions for two light sources to be coherent are as follows:
-The coherent sources of light must have originated from a single source and must be monochromatic (that is they have a single wavelength) in nature.
-The path difference between the light waves from the two sources must be small.

(b) Let, a monochromatic source of light with wavelength $\lambda$ is used to illuminate a narrow slit, $S$ therefore producing two coherent sources of light through the two slits $S_1$ and$S_2$.

Now, the light waves from the slits $S_1$ and$S_2$ superimpose with each other and reach point $P$ as shown in the figure and have a path difference of$S_{2}P-S_{1}P$.
Let $d=$ the separation between the slits, $D=$ the distance between the screen and the slits, $x=$ the distance between the centre of the screen to point $P$ and $\theta =$ the angle between $MP$ and $MO$. Now, draw a perpendicular,$S_{1}N$ on line $S_{2}P$ such that distances $P{S}_1$ and $PN$ are equal. Therefore, the path difference is$S_{2}N$. Let, the point $P$ be at a distance $x$ from $O$. Therefore,
$PE=x-{\displaystyle \frac{d}{2}}$ and $PF=x+{\displaystyle \frac{d}{2}}$ .

Consider the right-angled triangles in the figure, we get:
$$\Rightarrow {\left(S_{2}P\right)}^2-{\left(S_{1}P\right)}^2=\left[D^2+{\left(x+{\displaystyle \frac{d}{2}}\right)}^2\right]-\left[D^2+{\left(x-{\displaystyle \frac{d}{2}}\right)}^2\right]$$
Simplify the equation:
$\Rightarrow {\left(S_{2}P\right)}^2-{\left(S_{1}P\right)}^2=2xd$
$\Rightarrow \left(S_{2}P+S_{1}P\right)\left(S_{2}P-S_{1}P\right)=2xd$
$$\Rightarrow S_{2}P-S_{1}P={\displaystyle \frac{2xd}{S_{2}P+S_{1}P}}$$

Approximate that,
$\Rightarrow S_{2}P\cong S_{1}P\cong D$
So, Path difference $=S_{2}P-S_{1}P={\displaystyle \frac{xd}{D}}$
The condition for bright fringe at point P on the screen is:
$\Rightarrow S_{2}P-S_{1}P=m\lambda$
$\Rightarrow {\displaystyle \frac{xd}{D}}=m\lambda$
The condition for dark fringe at point P on the screen is:
$\Rightarrow S_{2}P-S_{1}P=\left(2m+1\right){\displaystyle \frac{\lambda }{2}}$
$\Rightarrow {\displaystyle \frac{xd}{D}}=\left(2m+1\right){\displaystyle \frac{\lambda }{2}}$
Where $m$ is the order of the fringe and $\lambda$ is the wavelength.

In order to find the fringe width, $\beta$ subtract the distance of the two consecutive bright or dark fringes.The fringe of $m^{th}$ order will be seen at:
$x_m={\displaystyle \frac{m\lambda D}{d}}$
The fringe of ${\left(m+1\right)}^{th}$ order will be seen at:
$x_{m+1}={\displaystyle \frac{\left(m+1\right)\lambda D}{d}}$
So, the fringe width,$\beta$ is given by:
$\Rightarrow \beta =x_{m+1}-x_m$
$\therefore \beta ={\displaystyle \frac{\lambda D}{d}}$

(c) If $s$ is the size of the source and $b$ its distance from the plane of the two slits, then the criterion for the interference fringes to occur is:
${\displaystyle \frac{s}{b}}<{\displaystyle \frac{\lambda }{d}}$
Where $d=$ the separation between the slits and $\lambda =$ wavelength of the monochromatic light.Also, the size of the source and the distance between the slits should be small.

Note: The fringe width for both the bright and dark fringes will be the same. This is because air is present between the slits and the screen and thus acting as a parallel film. Therefore, the light will not deviate from its original path and produce fringes of equal widths. If a Plano-convex lens is used to produce interference then circular fringes will occur as the air film will be of variable thickness.