Question

Unpolarized light passes through two polarizers in turn with polarization axes at \(45^{\circ}\) to one another.What is the fraction of the incident light intensity that is transmitted?

Short answer

Answer: \(\frac{1}{4}\)

Step-by-step solution

Recall Malus' law

Malus' law states that the intensity of the transmitted light through a polarizer can be calculated by the formula: \(I_t = I_i \cdot \cos^2{\theta}\), where \(I_t\) is the transmitted intensity, \(I_i\) is the initial intensity, and \(\theta\) is the angle between the polarization axis of the polarizer and the initial polarization direction.

Unpolarized light through the first polarizer

When unpolarized light passes through a polarizer, half the unpolarized light gets absorbed, and the light gets polarized with the remaining half intensity. So, after passing through the first polarizer, the intensity becomes \(I' = \frac{1}{2}I\), where \(I\) is the initial incident intensity and \(I'\) is the intensity after the first polarizer.

Polarized light through the second polarizer

Now the light is polarized and passes through the second polarizer with the axis at \(45^{\circ}\) to the first polarizer's axis. To find the transmitted intensity after the second polarizer, we use Malus' law: \(I_t = I' \cdot \cos^2{45^{\circ}}\) Recalling that \(\cos{45^{\circ}} = \frac{1}{\sqrt{2}}\), we substitute this into the equation: \(I_t = I' \cdot \left(\frac{1}{\sqrt{2}}\right)^2\)

Finding the fraction of transmitted intensity relative to initial intensity

Replace \(I'\) in the above equation with \(\frac{1}{2}I\): \(I_t = \left(\frac{1}{2}I\right) \cdot \left(\frac{1}{2}\right)\) \(I_t = \frac{1}{4}I\) To find the fraction of the transmitted intensity with respect to the initial intensity, divide \(I_t\) by \(I\): Fraction \(= \frac{I_t}{I} = \frac{\frac{1}{4}I}{I} = \frac{1}{4}\) So, the fraction of the incident light intensity that is transmitted through the two polarizers is \(\frac{1}{4}\).