Physics - Wave Optics Question with Solution | TestHub

Let $I_0$ be intensity of unpolarised light incident on first polaroid.
$I_1=$ Intensity of light transmitted from $1^{\text{st }}$ polaroid
$=\frac{I_0}{2}$
If $\theta$ be the angle between $1^{\text{st }}$ and $2^{\text{nd }}$ polaroid and $\varphi$ be the angle between $2^{\text{nd }}$ and $3^{\text{rd }}$ polaroid, then

$\theta +\varphi =90° ...\left(1\right)$

If $I_2=$ Intensity from $2^{\text{nd }}$ polaroid,

$\begin{array}{rcl}{I}_2& =& I_{1}co{s}^2\theta \\ & =& \frac{I_0}{2}co{s}^2\theta \end{array}$

If $I_3=$ Intensity from $3^{\text{rd }}$ polaroid

$\begin{array}{rcl}{I}_3& =& I_{2}co{s}^2\varphi \\ & =& I_1{\cos }^2\theta {\cos }^2\varphi \\ & =& \frac{I_0}{2}co{s}^2\theta co{s}^2\left(90°-\theta \right)\\ & =& \frac{I_0}{2}co{s}^2\theta si{n}^2\theta \\ & =& \frac{I_0}{2}{\left[\frac{2\sin \theta \cos \theta }{2}\right]}^2\\ & =& \frac{I_0}{8}{\sin }^{2}2\theta \end{array}$

Now, $I_3$ will be maximum when

$\sin 2\theta =1$

$\Rightarrow 2\theta =90°$

$\Rightarrow \theta =45°$