Question

Two polarizers are out of alignment by \(30.0^{\circ} .\) If light of intensity \(1.00 \mathrm{~W} / \mathrm{m}^{2}\) and initially polarized halfway between the polarizing angles of the two filters passes through both filters, what is the intensity of the transmitted light?

Short answer

Answer: The intensity of the transmitted light after passing through both polarizers is approximately \(0.933 \mathrm{~W} / \mathrm{m}^{2}\).

Step-by-step solution

Calculate the angle of initial light polarization relative to the first polarizer

As the light is initially polarized exactly halfway between the polarizing angles of two filters, the angle of polarization of the initial light relative to the first polarizer will be half of the angle between them: \(\theta_1 = \frac{30.0^{\circ}}{2} = 15.0^{\circ}\)

Calculate the intensity of light after passing through the first polarizer

Using Malus's Law, we can find the intensity of light after it passes through the first polarizer: \(I_1 = I_{initial} \cdot \cos^2(\theta_1)\) \(I_1 = (1.00 \mathrm{~W} / \mathrm{m}^{2}) \cdot \cos^2(15.0^{\circ})\) \(I_1 \approx 0.966 \mathrm{~W} / \mathrm{m}^{2}\)

Calculate the angle between the initial light polarization relative to the second polarizer

As the light has now passed through the first filter, the angle of polarization of initial light relative to the second polarizer will be the difference between the angle out of alignment and \(\theta_1\): \(\theta_2 = 30.0^{\circ} - 15.0^{\circ} = 15.0^{\circ}\)

Calculate the intensity of light after passing through the second polarizer

Using Malus's Law again, we can find the intensity of light after it passes through the second polarizer: \(I_{transmitted} = I_1 \cdot \cos^2(\theta_2)\) \(I_{transmitted} = (0.966 \mathrm{~W} / \mathrm{m}^{2}) \cdot \cos^2(15.0^{\circ})\) \(I_{transmitted} \approx 0.933 \mathrm{~W} / \mathrm{m}^{2}\) After passing through both polarizers, the intensity of the transmitted light is approximately \(0.933 \mathrm{~W} / \mathrm{m}^{2}\).