Question

To visually examine sunspots through a telescope, astronomers have to reduce the intensity of the sunlight to avoid harming their retinas. They accomplish this intensity reduction by mounting two polarizers on the telescope. The first polarizer has a polarizing angle of \(38.3^{\circ}\) relative to the horizontal. If the astronomers want to reduce the intensity of the sunlight by a factor of 0.7584 , what polarizing angle should the second polarizer have with the horizontal? Assume that this angle is greater than that of the first polarizer.

Short answer

Answer: Approximately 76 degrees.

Step-by-step solution

Write down the given information and formula

We know the following: - The polarizing angle of the first polarizer is \(38.3^\circ\). Let's call this angle \(\alpha\). - We want to reduce the intensity of the sunlight by a factor of 0.7584. Let's call this factor \(I\). Malus' law states that \(I = \cos^2 (\omega)\), where \(\omega\) is the angle between the polarizing axes of the two polarizers.

Calculate the angle between their polarizing axes for the required intensity

We want to find the angle \(\omega\) that corresponds to the intensity \(I = 0.7584\). Using Malus' law, we can write: \(0.7584 = \cos^2(\omega)\) To find \(\omega\), take the square root of 0.7584 and then take the inverse cosine: \(\omega = \arccos(\sqrt{0.7584})\) Calculate the value of \(\omega\): \(\omega \approx 37.7^\circ\)

Determine the polarizing angle of the second polarizer

Now we have found that the angle between the polarizing axes of the two polarizers is \(37.7^\circ\). We can use this information to find the polarizing angle of the second polarizer (\(\beta\)) relative to the horizontal. Since the angle \(\omega\) is the angle between the polarizing axes of the two polarizers and both axes are measured relative to the horizontal, we can write: \(\beta = \alpha + \omega\) Plug in the values of \(\alpha\) and \(\omega\): \(\beta = 38.3^\circ + 37.7^\circ\) \(\beta \approx 76^\circ\) The polarizing angle of the second polarizer relative to the horizontal should be approximately \(76^\circ\).