Young's Double Slit Experiment - Practice Questions & MCQ

Young's double-slit experiment -1-

This experiment is performed by British physicist Thomas Young. He used an arrangement as shown below. In this he used a monochromatic source of light S . He made two pinholes S₁ and S₂ (very close to each other) on an opaque screen as shown in the figure Each source can be considered as a source of the coherent light source. 

So, we can see that the monochromatic light source ‘s’ kept at a considerable distance from two slits s₁ and s₂. The arrangement is such that the S is equidistant from S₁ and S₂. S₁ and S₂ behave as two coherent sources, as both bring derived from S. 

Let d be the distance between two coherent sources A and B having wavelength λ. A screen XY is placed parallel to an opaque screen at a distance D. O is a point on the screen equidistant from A and B. P is a point at a distance x from O

From the above figure, we can see that the waves from A and B meet at P. It may be in phase or out of phase depending upon the path difference between the two waves, which we will calculate.

Draw AM perpendicular to BP
The path difference $\delta=\mathrm{BP}-\mathrm{AP}$
As we can see that, $\mathrm{AP}=\mathrm{MP}$

$
\delta=\mathrm{BP}-\mathrm{AP}=\mathrm{BP}-\mathrm{MP}=\mathrm{BM}
$

In right angled $\mathrm{ABM}, \quad \mathrm{BM}=\mathrm{d} \sin \theta$ lf $\theta$ is small,

$
\sin \theta=\theta
$

The path difference $\delta=\theta . d$
In right angled triangle $\mathrm{COP}, \quad \tan \theta=\mathrm{OP} / \mathrm{CO}=\mathrm{X} / \mathrm{D}$
For small values of $\theta, \tan \theta=\theta$
Thus, the path difference $\delta=\mathrm{xd} / \mathrm{D}$
So, the path differnece is $=\frac{x d}{D}$

For Bright Fringes - 

By the principle of interference, the condition for constructive interference is the path difference = nλ

$
\frac{x d}{D}=n \lambda
$

Here, $n=0,1,2 \ldots \ldots$ indicate the order of bright fringes
So, $x=\left(\frac{n \lambda D}{d}\right)$
This equation gives the distance of the $n^{\text {th }}$ bright fringe from the point O .

For Dark fringes :

By the principle of interference, the condition for destructive interference is the path difference $=$

$
\frac{(2 n-1) \lambda}{2}
$

Here, $n=1,2,3 \ldots$ indicate the order of the dark fringes.
So,

$
x=\frac{(2 n-1) \lambda D}{2 d}
$

The above equation gives the distance of the $\mathrm{n}^{\text {th }}$ dark fringe from the point O .
So, we can say that the alternate dark and bright fringe will be obtained on either side of the central bright fringe.

Band Width (β) -

The distance between any two consecutive bright or dark bands is called bandwidth. 

Take the consecutive dark or bright fringe - 

$
\begin{aligned}
x_{n+1}-x_n & =\frac{(n+1) \lambda D}{d}-\frac{(n) \lambda D}{d} \\
x_{n+1}-x_n & =\frac{\lambda D}{d} \\
\beta & =\frac{\lambda D}{d}
\end{aligned}
$

Angular fringe width -

$
\theta=\frac{\beta}{D}=\frac{\lambda D / d}{D}=\frac{\lambda}{d}
$
 

Young's double slit experiment- 2 -

Intensity of Fringes In Young’s Double Slit Experiment-

For two coherent sources S₁ and S₂, the resultant intensity at point P on the screen is given by-

$
I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi
$

where
$I_1=$ The intensity of wave from $\mathrm{S}_1$
$I_2=$ The intensity of wave $\mathrm{S}_2$
Putting $I_1$ and $I_2=I_o \quad$ (Becasue $\mathrm{d} \ll<\mathrm{D}$ )

$
\Rightarrow I=I_0+I_0+2 \sqrt{I_0 I_0} \cos \phi=4 I_0 \cos ^2 \frac{\phi}{2}
$

So the intensity variation from maximum to minimum depends on the phase difference. Let us discuss one by one -

For maximum intensity
The phase difference between the waves at the point of observation is $\phi=0^{\circ}$ or $2 n \pi$. Path difference between the waves at the point of observation is $\Delta x=n \lambda($ i.e. even multiple of $\lambda / 2$ )

Resultant intensity at the point of observation will be maximum

$
\text { i.e } \begin{aligned}
I_{\max } & =I_1+I_2+2 \sqrt{I_1 I_2} \\
I_{\max } & =\left(\sqrt{I_1}+\sqrt{I_2}\right)^2 \\
\text { If } \quad I_1 & =I_2=I_0 \Rightarrow I_{\max }=4 I_0
\end{aligned}
$

For Minimum Intensity -
The phase difference between the waves at the point of observation is

$
\begin{aligned}
& \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{aligned}
$
 

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)
$

Resultant intensity at the point of observation will be maximum

$
\begin{gathered}
I_{\min }=I_1+I_2-2 \sqrt{I_1 I_2} \\
\quad I_{\min }=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2 \\
\text { If } I_1=I_2=I_0 \Rightarrow I_{\min }=0
\end{gathered}
$
 

Maximum Order of Interference Fringes -

As we know that the position of n^th order maxima on the screen is - 

$
\frac{n \lambda D}{d} ; n=0, \pm 1, \pm 2, \ldots
$

Value of ' $n$ ' cannot be taken as infinitely large, because it violates the assumption of the Young's double slit experiment (Mentioned in the last concept) which means that the $\theta$ is small or we can write $\mathrm{x}<\mathrm{D}$. So,

$
\Rightarrow \frac{x}{D}=\frac{n \lambda}{d}<<1
$

So the above formula is only applicable for -

$
n \ll<\frac{d}{\lambda}
$

But when, $n \approx \frac{d}{\lambda}$, which means that the n is comparable with $\frac{d}{\lambda}$. Then the above formula is not applicable, then we have to go with the basic and we will equate path difference as -

$
\begin{aligned}
& \Rightarrow d \sin \theta=n \lambda \\
& \Rightarrow n=\frac{d \sin \theta}{\lambda} \\
& \text { So, } \\
& n_{\max }=\left[\frac{d}{\lambda}\right]
\end{aligned}
$

The above represents box function or greatest integer function.
Similarly, the highest order of interference minima

$
n_{\min }=\left[\frac{d}{\lambda}+\frac{1}{2}\right]
$