Interference Of Light - Condition And Types MCQ - Practice Questions with Answers

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 $(\Delta \phi)_{\text {: }}$
  The difference between the phases of two waves at a point is called phase difference.
  i.e. if $y_1=a_1 \sin \omega t$ and $y_2=a_2 \sin (\omega t+\phi)$ 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

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 $(\mathrm{y})$ which is the algebraic sum of the displacements ( $y_1$ and $y_2$ ) of each wave.

$$
\text { i.e } y=y_1+y_2
$$

consider two waves with the equations as

$$
\begin{aligned}
& y_1=A_1 \sin (k x-w t) \\
& y_2=A_2 \sin (k x-w t+\phi)
\end{aligned}
$$

where $\phi_{\text {is the phase difference between waves }} y_1$ and $y_2$.
And According to the principle of Super Position of waves

$$
\begin{aligned}
y= & y_1+y_2=A_1 \sin (k x-w t)+A_2 \sin (k x-w t+\phi) \\
& =A_1 \sin (k x-w t)+A_2[\sin (k x-w t) \cos \phi+\sin \phi \cos (k x-\omega t)] \\
\Rightarrow y & =\sin (k x-w t)\left[A_1+A_2 \cos \phi\right]+A_2 \sin \phi \cos (k x-w t) \ldots(1)
\end{aligned}
$$

Now let

$$
\begin{aligned}
& \quad A \cos \theta=A_1+A_2 \cos \phi \\
& \text { and } A \sin \theta=A_2 \sin \phi
\end{aligned}
$$

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 $A=\sqrt{A_1^2+A_2{ }^2+2 A_1 A_2 \cos \phi}$
and $\theta=\tan ^{-1}\left(\frac{A_2 \sin \phi}{A_1+A_2 \cos \phi}\right)$
where
$A_1=$ the amplitude of wave 1
$A_2=$ the amplitude of wave 2
- $A_{\max }=a_1+a_2$ and $A_{\text {min }}=a_1-a_2$

Resultant Intensity of two waves (I)-
Using $I \quad \alpha \quad A^2$
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

• . $I_{\max }=I_1+I_2+2 \sqrt{I_1 I_2} \Rightarrow I_{\max }=\left(\sqrt{I_1}+\sqrt{I_2}\right)^2$
  . $I_{\text {min }}=I_1+I_2-2 \sqrt{I_1 I_2} \Rightarrow I_{\text {min }}=\left(\sqrt{I_1}-\sqrt{I_2}\right)^2$
  - For identical source-

  $$
  I_1=I_2=I_0 \Rightarrow I=I_0+I_0+2 \sqrt{I_0 I_0} \cos \phi=4 I_0 \cos ^2 \frac{\phi}{2}
  $$

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

  $$
  \text { i.e } A_{\max }=a_1+a_2
  $$

  
  If $a_1=a_2=a_0 \Rightarrow A_{\max }=2 a_0$
  - Resultant intensity at the point of observation will be maximum

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

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 

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

- Path difference between the waves at the point of observation is

$$
\Delta x=(2 n-1) \frac{\lambda}{2}(\text { i.e. odd multiple of } \lambda / 2)
$$

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

$$
\text { i.e } A_{\min }=a_1-a_2
$$

If $a_1=a_2 \Rightarrow A_{\text {min }}=0$
- Resultant intensity at the point of observation will be maximum

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