Question

Unpolarized light of intensity \(I_{0}\) is incident on a series of five polarizers, with the polarization direction of each rotated \(10.0^{\circ}\) from that of the preceding one. What fraction of the incident light will pass through the series?

Short answer

Answer: About 0.82% of the incident light will pass through the series of five polarizers.

Step-by-step solution

Intensity after the first polarizer

Since the incident light is unpolarized, half of its intensity is aligned with the polarizing direction of the first polarizer, which can be represented as follows: $$I_1 = \frac{1}{2}I_0$$

Intensity after the second polarizer

The angle between the polarizing directions of the first and second polarizers is \(10.0^{\circ}\). Using Malus's law: $$I_2 = I_1\cos^2(10.0^{\circ})$$ Substitute the value of \(I_1\) from Step 1: $$I_2 = \frac{1}{2}I_0\cos^2(10.0^{\circ})$$

Intensity after the third polarizer

The angle between the polarizing directions of the second and third polarizers is again \(10.0^{\circ}\). Using Malus's law: $$I_3 = I_2\cos^2(10.0^{\circ})$$ Substitute the value of \(I_2\) from Step 2: $$I_3 = \frac{1}{2}I_0\cos^2(10.0^{\circ})\cos^2(10.0^{\circ})$$

Intensity after the fourth polarizer

The angle between the polarizing directions of the third and fourth polarizers is \(10.0^{\circ}\). Using Malus's law: $$I_4 = I_3\cos^2(10.0^{\circ})$$ Substitute the value of \(I_3\) from Step 3: $$I_4 = \frac{1}{2}I_0\cos^2(10.0^{\circ})\cos^2(10.0^{\circ})\cos^2(10.0^{\circ})$$

Intensity after the fifth polarizer

The angle between the polarizing directions of the fourth and fifth polarizers is \(10.0^{\circ}\). Using Malus's law: $$I_5 = I_4\cos^2(10.0^{\circ})$$ Substitute the value of \(I_4\) from Step 4: $$I_5 = \frac{1}{2}I_0\cos^2(10.0^{\circ})\cos^2(10.0^{\circ})\cos^2(10.0^{\circ})\cos^2(10.0^{\circ})$$

Calculate the fraction of transmitted light

To find the fraction of transmitted light, divide the final intensity \(I_5\) by the initial intensity \(I_0\): $$f = \frac{I_5}{I_0}$$ Substitute the value of \(I_5\) from Step 5: $$f = \frac{1}{2}\cos^2(10.0^{\circ})\cos^2(10.0^{\circ})\cos^2(10.0^{\circ})\cos^2(10.0^{\circ})$$ The fraction of the incident light that will pass through all five polarizers is: $$f ≈ 0.0082$$ So, about 0.82% of the incident light will pass through the series of five polarizers.