Question

Vertically polarized light with intensity \(I_{0}\) is normally incident on an ideal polarizer. As the polarizer is rotated about a horizontal axis, the intensity \(I\) of light transmitted through the polarizer varies with the orientation of the polarizer \((\theta),\) where \(\theta=0\) corresponds to a vertical transmission axis. Sketch a graph of \(I\) as a function of \(\theta\) for one complete rotation of the polarizer $\left(0 \leq \theta \leq 360^{\circ}\right)$.(W) tutorial: polarized light).

Short answer

Answer: The graph of the transmitted light intensity (I) as a function of the orientation of the polarizer (θ) according to Malus' law shows a cosine wave behavior for one complete rotation (0 ≤ θ ≤ 360°). It has two maxima at θ = 0° and 180°, and two minima at θ = 90° and 270°.

Step-by-step solution

Define the function relating the transmitted light intensity and the orientation of the polarizer

According to the Malus' law, we have I = I0 * cos²(θ).

Plot the graph of I as a function of θ

To sketch the graph of I as a function of θ, we need to find the values of I at different values of θ. Let's find the values of I at θ = 0°, 45°, 90°, 135°, 180°, 225°, 270°, and 315°. - At θ = 0°, I = I0 * cos²(0) = I0. - At θ = 45°, I = I0 * cos²(45) = I0 * (1/2). - At θ = 90°, I = I0 * cos²(90) = 0. - At θ = 135°, I = I0 * cos²(135) = I0 * (1/2). - At θ = 180°, I = I0 * cos²(180) = I0. - At θ = 225°, I = I0 * cos²(225) = I0 * (1/2). - At θ = 270°, I = I0 * cos²(270) = 0. - At θ = 315°, I = I0 * cos²(315) = I0 * (1/2). Now we can sketch the graph based on the above values. In the graph, the x-axis represents θ, while the y-axis represents I. The graph should be a continuous curve starting at I0 when θ = 0°, decreasing to 0 at θ = 90°, then rising back to I0 at θ = 180°, then falling back to 0 at θ = 270°, and finally rising back to I0 at θ = 360°. The graph will show a cosine wave behavior, with two maxima at θ = 0° and 180° and two minima at θ = 90° and 270°.