Light polarized in the \(x\) -direction shines through two polarizing sheets. The first sheet's transmission axis makes an angle \(\theta\) with the \(x\) -axis and the transmission axis of the second is parallel to the \(y\) -axis. (a) If the incident light has intensity \(I_{0},\) what is the intensity of the light transmitted through the second sheet? (b) At what angle \(\theta\) is the transmitted intensity maximum?

Since the initial light is polarized in the x-direction and makes an angle \(\theta\) with the first sheet's transmission axis, applying Malus's law to calculate the light intensity, \(I_{1}\), after passing through the first sheet: \(I_{1} = I_{0}\cos^2{\theta}\)

Now, the transmitted light through the first sheet has the polarization plane making an angle \(\theta\) with the x-axis. The transmission axis of the second sheet is parallel to the y-axis. Therefore, the angle between the polarization plane and the transmission axis of the second sheet is equal to \(90^{\circ} - \theta\). Applying Malus's law again, we can find the intensity, \(I_{2}\), after passing through the second sheet: \(I_{2} = I_{1}\cos^2{(90^{\circ}-\theta)}\) Using our result from Step 1, we have \(I_{2} = I_{0}\cos^2{\theta}\cos^2{(90^{\circ}-\theta)}\) (a) The intensity of the light transmitted through the second sheet is: \(I_{2} = I_{0}\cos^2{\theta}\cos^2{(90^{\circ}-\theta)}\)

To find the angle at which the transmitted intensity is maximum, we differentiate \(I_{2}\) with respect to \(\theta\) and set the derivative to 0: \(\frac{dI_{2}}{d\theta} = 0\) Taking the derivative, we have: \(\frac{dI_{2}}{d\theta} = I_{0}\left[ -2\cos{\theta}\sin{\theta}\cos^2{(90^{\circ} - \theta)} - \sin^2{\theta}\cos{(90^{\circ}-\theta)}\sin{(90^{\circ}-\theta)} \right]\) Setting the derivative to 0, we get: \(-2\cos{\theta}\sin{\theta}\cos^2{(90^{\circ} - \theta)} - \sin^2{\theta}\cos{(90^{\circ}-\theta)}\sin{(90^{\circ}-\theta)} = 0\) This equation simplifies to: \(\cos{\theta}\sin{\theta}\left[ -2\cos^2{(90^{\circ} - \theta)} - \sin^2{\theta} \right] = 0\) The angle \(\theta\) must be between 0 and 90 degrees. So, the maximum intensity occurs when: \(\cos{\theta}\sin{\theta} = 0\) The only solution for \(\theta\) in this case is \(0^{\circ}\) or \(90^{\circ}\). Since we are given that the x-axis polarization is non-zero, the maximum transmitted intensity will occur when the angle \(\theta = 0^{\circ}\). (b) The angle at which the transmitted intensity is maximum is \(\theta = 0^{\circ}\).