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Wave Optics — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsWave Optics5 Marks
Question


Figure shown a two slit arrangement with a source which emits unpolarised light. $P$ is a polariser with axis whose direction is not given. If $I_0$ is the intensity of the principal maxima when no polariser is present, calculate in the present case, the intensity of the principal maxima as well as of the first minima.
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Answer

The resultant amplitude of wave reaching on screen will be the sum of amplitude of either wave in perpendicular and parallel polarisation.
Amplitude of the wave in perpendicular polarisation
$\text{A}_\bot=\text{A}_\bot^1+\text{A}_\bot^2=\text{A}_\bot^0\sin(\text{kx}-\omega\text{t})+\text{A}_\bot^2\sin(\text{kx}-\omega\text{t}+\phi)$
$\Rightarrow\ \text{A}_\bot=\text{A}_\bot^0(\sin(\text{kx}-\omega\text{t})+\sin(\text{kx}-\omega\text{t}+\phi))$
Amplitude of the wave in parallel polarisation
$\text{A}_{||}=\text{A}_{||}^{(1)}+\text{A}_{||}^{(2)}$
$\Rightarrow\ \text{A}_\bot=\text{A}_\bot^0[\sin(\text{kx}-\omega\text{t})+\sin(\text{kx}-\omega\text{t}+\phi)]$
$\therefore$ Average Intensity on the screen
$\text{I}=\begin{Bmatrix} \Big|\text{A}_\bot^0\Big|^2+\Big|\text{A}_\bot^0\Big|^2\end{Bmatrix}\big[\sin(\text{kx}-\omega\text{t})(1+\cos^2\phi+2\sin\phi)+\sin^2(\text{kx}-\omega\text{t})\sin^2\phi\big]_\text{average}$
$=\begin{Bmatrix} \Big|\text{A}_\bot^0\Big|^2+\Big|\text{A}_\bot^0\Big|^2\end{Bmatrix}\Big(\frac{1}{2}\Big)2(1+\cos\phi)$
$\Rightarrow\ \text{I}=2\Big|\text{A}_{||}^0\Big|^2(1+\cos\phi)\text{ since},\Big|\text{A}_\bot^0\Big|_\text{av}+\Big|\text{A}_\bot^0\Big|_\text{av}$
With polariser $P,$
Assume $\text{A}_\bot^2$ is blocked
Intensity $=(\text{A}_{||}^1+\text{A}_{||}^2)^2+({\text{A}_{||}^2})^2$
$=\Big|\text{A}_{\bot}^0\Big|(1+\cos\phi)+\Big|\text{A}_{\bot}^0\Big|.\frac{1}{2}$
Given, $\text{I}_0=4\Big|\text{A}_{\bot}^0\Big|=$ Intensity without polariser at principal maxima.
Intensity at first minima with polariser
$=\Big|\text{A}_{\bot}^0\Big|(1-1)+\frac{\Big|\text{A}_{\bot}^0\Big|}{2}=\frac{\text{I}_0}{8}.$

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