# A vertically polarized light that has an intensity of 430 W/m^2 is incident on two polarizing...

## Question:

A vertically polarized light that has an intensity of {eq}\rm 430 \ W/m^2 {/eq} is incident on two polarizing filters. The first filter is oriented 30 degrees from the vertical while the second filter is oriented 75 degrees from the vertical. Predict the intensity and polarization of the light that emerges from the second filter. (Assume both angles are measured in the same direction with respect to the vertical.)

## Polarisation And Malus's Law

Polarization (also polarisation) is a property applying to transverse waves that specifies the geometrical orientation of the oscillations.

An electromagnetic wave such as light consists of a coupled oscillating electric field and magnetic field which are always perpendicular; by convention, the "polarization" of electromagnetic waves refers to the direction of the electric field. In linear polarization, the fields oscillate in a single direction. In circular or elliptical polarization, the fields rotate at a constant rate in a plane as the wave travels. The rotation can have two possible directions; if the fields rotate in a right hand sense with respect to the direction of wave travel, it is called right circular polarization, or, if the fields rotate in a left hand sense, it is called left circular polarization.

According to malus, when completely plane polarized light is incident on the analyzer, the intensity I of the light transmitted by the analyzer is directly proportional to the square of the cosine of angle between the transmission axes of the analyzer and the polarizer.

{eq}I\ \propto\ (Cos{\theta})^2 {/eq}

then,

{eq}I\ =\ I_o\ \times\ (Cos{\theta})^2 {/eq}

where:

• {eq}I {/eq} is the intensity of the light crossed the polarizer
• {eq}I_o {/eq} is the original intensity of the polarised light before crossing the polarizer
• {eq}\theta {/eq} is the polarization angle.

Become a Study.com member to unlock this answer! Create your account

Using the above law,

Here the intensity after passing from the first polarizer is equal to {eq}=\ 430\ (\ Cos30\ )^2\ W/m^2\ =\ 322.5\ W/m^2 {/eq}

... 