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)

$\frac{I_0}{8}$

Without $P$;

$A=A_{\perp }+A_{11}$

$A_1=A_{\perp }^1+A_{\perp }^2=A_{\perp }^0\sin (kx-wt)+A_{\perp }^0\sin (kx-wt+\varphi )$

$A_{11}=A_{11}^{\left(1\right)}+A_{11}^{\left(2\right)}$

$A_{11}=A_{11}^0\left[\sin \right(kx-wt)+\sin (kx-wt+\varphi )$

where $A_{\perp }^0,A_{11}^0$

are the amplitudes of either of the beam in $\perp$ and 11 polarizations.

$\therefore$ Intensity $=\{|A_{\perp }^0{|}^2+|A_{11}^0{|}^2\}\left({\sin }^2\right(kx-wt)$

$(1+co{s}^2\varphi +2\sin \varphi )+{\sin }^2(kx-wt){\sin }^2\varphi {)}_{average}$

$=\{|A_{\perp }^0{|}^2+|A_{11}^0{|}^2\}\left(\frac{1}{2}\right).2(1+\cos \varphi )$

$=2|A_{\perp }^0{|}^2.(1+\cos \varphi )since|A_{\perp }^0{|}_{average}=|A_{11}^0{|}_{average}$

With $P$:

Assume $A_{\perp }^2$ is blocked:

Intensity $=(A_{11}^1+A_{11}^1{)}^2+(A_{\perp }^1{)}^2$

$=|A_{\perp }^0{|}^2(1+\cos \varphi )+|A_{\perp }^0{|}^2{.}_2^1$

Given: $I_0=4|A_{bot}^0{|}^2$

$=$ Intensity wothout polariser at principal maxima.

intensity at principal maxima with polariser

$|A_{\perp }^0{|}^2(2+\frac{1}{2})$

$=\frac{5}{8}{I}_0$

Intensity at first minima with polariser

$=|A_{\perp }^0{|}^2(1-1)+\frac{|A_1^0{|}^2}{2}$

$=\frac{I_0}{8}$.