A clear sheet of polaroid is placed on the top of a similar sheet so that their axes make an angle ${\sin }^{-1}\left(\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)$ with each other. The ratio of intensity of the emergent light to that of unpolarized light is

Let ${\mathrm{I}}_{\mathrm{o}}$ be the intensity of unpolarized light.

The intensity transmitted by the first sheet is ${\mathrm{I}}_1=\raisebox{1ex}{${\mathrm{I}}_{\mathrm{o}}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$.

This will be the intensity of incident light on the second polaroid. The intensity transmitted by the second polaroid will be 
$$\begin{gathered} I_2=\left(\raisebox{1ex}{${\mathrm{I}}_0$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right){\cos }^2\mathrm{\theta } \\ , \text{ where }\mathrm{\theta }\text{ is the angle between the axes } \end{gathered}$$

$\because \sin \mathrm{\theta }=3/5 \Rightarrow \cos \mathrm{\theta }=(4/5)$

$\Rightarrow {\mathrm{I}}_2= \frac{{\mathrm{I}}_{\mathrm{o}}}{2}\times {\left(\frac{4}{5}\right)}^2=\frac{8}{25}{\mathrm{I}}_{\mathrm{o}}$

$$\begin{gathered} \mathrm{so}, \mathrm{Ratio} \mathrm{of} \mathrm{intensity} \mathrm{of} \mathrm{emergent} \mathrm{light} \mathrm{to} \mathrm{that} \mathrm{of} \mathrm{unpolarised} \mathrm{light} \\ \frac{{\mathrm{I}}_2}{{\mathrm{I}}_{\mathrm{o}}}=\frac{8}{25} \end{gathered}$$