Light cannot be transmitted through two coaxial polarising filters if their polarisation axes are or

Question

Light cannot be transmitted through two coaxial polarising filters if their polarisation axes are orthogonally oriented. However, if a third filter is placed between them, some light can pass through all three. An unpolarized beam of light of intensity $I_{0}$ is incident on the first filter. Let $f_{max}$ be the maximum possible fraction of the incident unpolarized intensity that can be transmitted through this three-filter system. 
Find the value of $(100 	imes f_{max})$.

Professor PC Thomas Classes · Answer

Let the initial unpolarized intensity be $I_{0}$. After passing through the first filter, the intensity becomes $I_{1} = I_{0} / 2$, and the light becomes plane-polarized. Let the middle filter be oriented at an angle $\phi$ relative to the first filter's transmission axis. According to Malus's Law, the intensity after the second filter is $I_{2} = I_{1} \cos^{2}(\phi) = (I_{0} / 2) \cos^{2}(\phi)$. The third filter is orthogonal to the first, meaning its angle relative to the second filter is $(90^{\circ} - \phi)$. The intensity after the third filter is $I_{3} = I_{2} \cos^{2}(90^{\circ} - \phi) = I_{2} \sin^{2}(\phi)$. Substitute $I_{2}$: $I_{3} = (I_{0} / 2) \cos^{2}(\phi) \sin^{2}(\phi)$. Using the trigonometric identity $\sin(2\phi) = 2 \sin(\phi) \cos(\phi)$, we get $I_{3} = (I_{0} / 8) \sin^{2}(2\phi)$. The maximum transmitted intensity occurs when $\sin^{2}(2\phi) = 1$, which happens at $\phi = 45^{\circ}$. Maximum intensity $= I_{0} / 8$. The fraction $f_{max} = 1/8 = 0.125$. Therefore, $100 	imes f_{max} = 12.5$.