ns. Diagram production of polarized light by scattering of sun light
● Explanation
● Calculation of intensity of light transmitted through ${\mathit{P}}_1,{\mathit{P}}_2$ and ${\mathit{P}}_3$
Diagram :
Explanation: Charges accelerating parallel to the double arrows do not radiate energy towards the observer. The radiation scattered by the molecules therefore is polarised perpendicular to the plane of the figure.
Alternatively: If the student writes " scattered light when viewed in a perpendicular direction is found to be polarised
(Award one mark) Intensity of light transmitted by Polaroid is,
${\mathit{I}}_1=\frac{\mathit{I}}{2}$
Intensity of light transmitted by Polaroid is,
${\mathit{I}}_2={\mathit{I}}_1{\cos }^2{45}^{\circ }$ $=\frac{\mathit{I}}{2}{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{\mathit{I}}{4}$
Intensity of light transmitted by Polaroid is,
${\mathit{I}}_3={\mathit{I}}_1{\cos }^2{45}^{\circ }$ $=\frac{\mathit{I}}{2}{\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{\mathit{I}}{8}$
OR

● Reason
● Deriving the expression for resultant intensity and condition for constructive and destructive interference
● Calculating the separation
(a) Because two independent sources cannot be coherent OR they are not coherent
(b) ${\mathit{y}}_1=\mathit{a}\cos \mathit{\omega }\mathit{t}$
${\mathit{y}}_2=\mathit{a}\cos \left(\mathit{\omega }\mathit{t}+\mathit{\varphi }\right)$
So resultant displacement is give by
$\mathit{y}={\mathit{y}}_1+{\mathit{y}}_2$
$\mathit{y}=\mathit{a}\cos \mathit{\omega }\mathit{t}+\mathit{a}\cos \left(\mathit{\omega }\mathit{t}+\mathit{\varphi }\right)$
$\mathit{y}=2\mathit{a}\cos \left(\mathit{\varphi }∕2\right)\cos \left(\mathit{\omega }\mathit{t}+\mathit{\varphi }∕2\right)$
The amplitude of the resultant displacement is $2\mathit{\pi }\cos \left(\mathit{\varphi }∕2\right)$ and therefore intensity at that point will be $\mathit{I}=4{\mathit{I}}_0{\cos }^2\left(\mathit{\varphi }∕2\right)$
For constructive interference:
$\mathit{\varphi }=0,\pm 2\mathit{\pi },\pm 4\mathit{\pi },.......$
For destructive interference:
$\mathit{\varphi }=0,\pm \mathit{\pi },\pm 3\mathit{\pi },\pm 5\mathit{\pi },.......$
(c) Position of second maxima,
${\mathit{y}}_2=\frac{5\mathit{\lambda }}{2}\frac{\mathit{\lambda }\mathit{D}}{\mathit{a}}$
Separation between the positions of the second maxima with ${\mathit{\lambda }}_1$ and ${\mathit{\lambda }}_2$ is:
$∆\mathit{y}=\frac{5\mathit{D}\left({\mathit{\lambda }}_2-{\mathit{\lambda }}_1\right)}{2\mathit{a}}$
$=\frac{5\times 1.5\times \left(596-590\right)\times {10}^{-9}}{2\times 2\times {10}^{-6}}$
$=11.25\times {10}^{-3}\mathrm{m}$