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Chapter 22: Problem 70

Polarized light of intensity \(I_{0}\) is incident on a pair of polarizing sheets. Let \(\theta_{1}\) and \(\theta_{2}\) be the angles between the direction of polarization of the incident light and the transmission axes of the first and second sheets, respectively. Show that the intensity of the transmitted light is $I=I_{0} \cos ^{2} \theta_{1} \cos ^{2}\left(\theta_{1}-\theta_{2}\right)$.

Short Answer

Expert verified
Answer: The intensity of the transmitted light can be found using the formula I = I鈧� * cos虏(胃鈧�) * cos虏(胃鈧� - 胃鈧�).

Step by step solution

01

Intensity after the first sheet

To find the intensity after the first polarizing sheet, apply Malus' Law using the angle between the direction of polarization of the incident light (胃鈧�) and the transmission axis of the first sheet. Since the initial intensity is I鈧�, the intensity after the first sheet (I鈧�) is: I鈧� = I鈧� * cos虏(胃鈧�)
02

Intensity after the second sheet

The intensity after the second sheet depends on both the first and the second sheet's transmission axes. Consider the angle between the first sheet's transmission axis (aligned with the transmitted light after the first sheet) and the second sheet's transmission axis, which is (胃鈧� - 胃鈧�). Now we apply Malus' Law again using this angle and the intensity after the first sheet (I鈧�): I = I鈧� * cos虏(胃鈧� - 胃鈧�)
03

Combining the results from both sheets

Now, substitute the expression for I鈧� obtained in step 1 into the equation we derived in step 2: I = (I鈧� * cos虏(胃鈧�)) * cos虏(胃鈧� - 胃鈧�) So, the intensity of the transmitted light is given as: I = I鈧� * cos虏(胃鈧�) * cos虏(胃鈧� - 胃鈧�)

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