Question

As shown in figure, 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.

• A. $\frac{I_0}{8}$
• B. $\frac{3}{4}{I}_0$
• C. $\frac{I_0}{16}$
• D. $\frac{2}{5}{I}_0$

Answer 1

Answer: (A)

Without $P$:
$A=A_{\perp }+A_{||}$.
$A_{\perp }=A_{\perp }^1+A_{\perp }^2=A_{\perp }^{0}sin\left(kx-\omega t\right)+A_{\perp }^{0}sin\left(kx-\omega t+\varphi \right)$
$A_{||}$$=A_{||}^1+A_{||}^2=A_{||}^0\left[sin\left(kx-\omega t\right)+sin\left(kx-\omega t+\varphi \right)\right]$
where $A_{\perp }^0,A_{||}^0$ are the amplitudes of either of the beam in $\perp$ and $||$ polarizations.
$\therefore$ Intensity $=\left\{{\left|A_{\perp }^0\right|}^2+{\left|A_{||}^0\right|}^2\right\}[si{n}^2\left(kx-\omega t\right)$
$\left(1+co{s}^2\varphi +2sin\varphi \right)+si{n}^2\left(kx-\omega t\right)si{n}^2\varphi {]}_{\text{a}verage}$
$=\left\{{\left|A_{\perp }^0\right|}^2+{\left|A_{||}^0\right|}^2\right\}\left(\frac{1}{2}\right)\cdot 2\left(1+cos\varphi \right)$
$=2{\left|A_{\perp }^0\right|}^2\left(1+cos\varphi \right),since{\left|A_{\perp }^0\right|}_{average}={\left|A_{||}^0\right|}_{average}$
With $P:$
Assume $A_{\perp }^2$ is blocked :
Intensity $={\left(A_{||}^1+A_{||}^2\right)}^2+{\left(A_{\perp }^1\right)}^2$
$={\left|A_{\perp }^0\right|}^2\left(1+cos\varphi \right)+{\left|A_{\perp }^0\right|}^2.\frac{1}{2}$
$(I_0=4{\left|A_{\perp }^0\right|}^2$= Intensity without polariser at principal maxima )
Intensity at first maxima with polariser
$={\left|A_{\perp }^0\right|}^2\left(2+\frac{1}{2}\right)$
$=\frac{5}{8}{I}_0$
Intensity at first minima with polariser,
${\left|A_{\perp }^0\right|}^2\left(1-1\right)+\frac{{|A_{\perp }^0|}^2}{2}$
$=\frac{I_0}{8}$