Question

A vertically polarized electromagnetic wave passes through the five polarizers in FIGURE$\mathrm{Q}31.10$. Rank in order, from largest to smallest, the transmitted intensities${\mathrm{I}}_{\mathrm{a}}$to${\mathrm{I}}_{\mathrm{e}}$.

Short answer

A vertically polarised electromagnetic wave goes through the five polarizers, which are ranked from largest to smallest in order of size is${\mathrm{I}}_{\mathrm{d}}>{\mathrm{I}}_{\mathrm{c}}={\mathrm{I}}_{\mathrm{e}}>{\mathrm{I}}_{\mathrm{b}}>{\mathrm{I}}_{\mathrm{a}}$.

Step-by-step solution

Figure for polarizes

Figure for The five polarizers pass a vertically polarized electromagnetic pulse,

Explanation

The amount of light that passes through a polarizer is${\mathrm{I}}_0{\cos }^2\left(\mathrm{\theta }\right)$.

The incident wave$0$is aligned with the polarizer$\mathrm{d}$.

while$\mathrm{\theta }=90°$for polarizer a.

Polarizers$\mathrm{c}$and$\mathrm{d}$are perpendicular to the vertical.

So,

${\mathrm{I}}_{\mathrm{d}}>{\mathrm{I}}_{\mathrm{c}}={\mathrm{I}}_{\mathrm{e}}>{\mathrm{I}}_{\mathrm{b}}>{\mathrm{I}}_{\mathrm{a}}$