Question

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?

Short answer

In summary, the intensity of light transmitted through the second sheet is given by: \(I_{2} = I_{0}\cos^2{\theta}\cos^2{(90^{\circ}−\theta)}\) And the angle at which the transmitted intensity is maximum is \(\theta = 0^{\circ}\).

Step-by-step solution

Compute intensity after passing through the first sheet

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: \(I_{1} = I_{0}\cos^2{\theta}\)

Compute intensity after passing through the second sheet

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: \(I_{2} = I_{0}\cos^2{\theta}\cos^2{(90^{\circ}-\theta)}\)

Find the angle at which the transmitted intensity is maximum

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 \(\theta = 0^{\circ}\).