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Light Waves — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsLight Waves5 Marks
Question
Consider the arrangement shown in figure. By some mechanism, the separation between the slits $S_3$ and $S_4$ can be changed. The intensity is measured at the point $P$ which is at the common perpendicular bisector

of $S_1S_2$ and $S_3S_4$. When $\text{z}=\frac{\text{D}\lambda}{2\text{d}},$ intensity measured at $P$ is I. Find this intensity when $z$ is equal to:
  1. $\frac{\text{D}\lambda}{\text{d}}$
  2. $\frac{3\text{D}\lambda}{2\text{d}}$
  3. $\frac{2\text{D}\lambda}{\text{d}}$
✓

Answer


  1. When, $\text{z}=\frac{\text{D}\lambda}{\text{d}}$
So, $\text{OS}_3=\text{OS}_4=\frac{\text{D}\lambda}{2\text{d}}$
$\Rightarrow$ Dark fringe at $S_3$ and $S_4.$
$\Rightarrow$ At $S_3$, intensity at $S_3 = 0$
$\Rightarrow I_1 = 0$
At $S_4,$ intensity at $S_4 = 0$
$\Rightarrow I_2 = 0$
At P, path difference $= 0$
$ \Rightarrow$ Phase difference $= 0.$
$\Rightarrow\text{I}=\text{I}_1+\text{I}_2+\sqrt{\text{I}_1\text{I}_2}\cos0^\circ=0+0+0=0$
$\Rightarrow$ Intensity at $P = 0.$
  1. Given that, when $\text{z}=\frac{\text{D}\lambda}{2\text{d}}$, intensity at $P = I$
Here, $\text{OS}_3=\text{OS}_4=\text{y}=\frac{\text{D}\lambda}{4\text{d}}$
$\therefore\phi=\frac{2\pi\text{x}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{\text{D}\lambda}{4\text{d}}\times\frac{\text{d}}{\text{D}}=\frac{\pi}{2}.$
$\Big[$Since, $x =$ path difference $=\frac{\text{yd}}{\text{D}}\Big]$
Let, intensity at $S_3$ and $S_4 = I'$
$\therefore$ At $P$, phase difference $= 0$
So, $\text{I}'+\text{I}'+2\text{I}'\cos0^\circ=\text{I}$
$\Rightarrow4\text{I}'=\text{I}\Rightarrow\text{I}'=\frac{1}{4}$
When, $\text{z}=\frac{3\text{D}\lambda}{2\text{d}},\Rightarrow\text{y}=\frac{3\text{D}\lambda}{4\text{d}}$
$\therefore\phi=\frac{2\pi\text{x}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{3\text{D}\lambda}{4\text{d}}\times\frac{\text{d}}{\text{D}}=\frac{3\pi}{2}$
Let,$I$'' be the intensity at $S_3$ and $S_4$ when, $\phi=\frac{3\pi}{2}$
Now comparing,
$\frac{\text{I}"}{\text{I}}=\frac{\text{a}^2+\text{a}^2+2\text{a}^2\cos\Big(\frac{3\pi}{2}\Big)}{\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}}=\frac{2\text{a}^2}{2\text{a}^2}=1$ $\Rightarrow\text{I}"=\text{I}'=\frac{\text{I}}{4}$
$\therefore$ Intensity at $\text{P}=\frac{\text{I}}{4}+\frac{\text{I}}{4}+2\times\Big(\frac{\text{I}}{4}\Big)\cos0^\circ=\frac{\text{I}}{2}+\frac{\text{I}}{2}=1$
  1. When $\text{z}=\frac{2\text{D}\lambda}{\text{d}}$
$\Rightarrow\text{y}=\text{OS}_3=\text{OS}_4=\frac{\text{D}\lambda}{\text{d} }$
$\therefore\phi=\frac{2\pi\text{X}}{\lambda}=\frac{2\pi}{\lambda}\times\frac{\text{yd}}{\text{D}}=\frac{2\pi}{\lambda}\times\frac{\text{D}\lambda}{\text{d}}\times\frac{\text{d}}{\text{D}}=2\pi.$
Let, $I"' =$ intensity at $S_3$ and $S_4$ when, $\phi=2\pi.$
$\frac{\text{I}'''}{\text{I}'}=\frac{\text{a}^2+\text{a}^2+2\text{a}^2\cos2\pi}{\text{a}^2+\text{a}^2+2\text{a}^2\cos\frac{\pi}{2}}=\frac{4\text{a}^2}{2\text{a}^2}=2$
$\Rightarrow\text{I}'''=2\text{I}'=2\Big(\frac{\text{I}}{4}\Big)=\frac{\text{I}}{2}$
At $P, I_{resultant} =\frac{\text{I}}{2}+\frac{\text{I}}{2}+2\Big(\frac{\text{I}}{2}\Big)\cos0^\circ=\text{I}+\text{I}=2\text{I}$
So, the resultant intensity at $P$ will be $2I.$

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