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Wave Optics — Physics STD 12 Science — Question

Bihar 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 I0 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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