Question

To practice Problem-Solving Strategy 33.2 Linear Polarization. Unpolarized light of intensity 30 W/cm2 is incident on a linear polarizer set at the polarizing angle θ1 = 14 ∘. The emerging light then passes through a second polarizer that is set at the polarizing angle θ2 = 162 ∘. Note that both polarizing angles are measured from the vertical. What is the intensity of the light that emerges from the second polarizer?

Answer 1

To solve this problem it is necessary to apply the concepts related to the law of Malus.

The light intensity after passing through the first polarizer is given as

$I_1 = \frac{I}{2}$

Where,

I = Intensity of unpolarized light

The expression for malus law is given as

$I_2 = I_1 Cos^2 \theta$

Where

\theta = Angle between the first and second polarizers

$I_1,I_2$  = Intensities of the light after passing through the first and second polarizers respectively.

The intensity of light after passing first polarizer ( Perpendicular to the surface) is

$I_1 = \frac{I}{2}$

$I_1 = \frac{30}{2}$

$I_1 = 15W/cm^2$

The angle between the first and second polarizers is given by

$\theta = \theta_2 -\theta_1$

$\theta = 162-14$

$\theta =148\°$

The expression for Malus law is given by

$I_2 = I_1 cos^2 \theta$

$\theta = 148\°$

$I_2 = (15)cos^2(148)$

$I_2 = 10.78W/cm^2$

Therefore the intensity of the light after passing through both the polarizer is $10.78W/cm^2$