If a polarizing filter reduces the intensity of polarized light to \(50.0 \%\) of its original value, by how much are the electric and magnetic fields reduced?

The intensity (I) of polarized light is directly proportional to the square of the electric field (E) and magnetic field (B). The formula for intensity is given by: \( I \propto E^2 \) and \( I \propto B^2 \) Since we want to find the new value of electric and magnetic fields, let's represent the initial intensity as \( I_0 \), final intensity as \( I_f \), initial electric field as \( E_0 \), final electric field as \( E_f \), initial magnetic field as \( B_0 \), and final magnetic field as \( B_f \).

The polarizing filter reduces the intensity of polarized light to 50.0% of its original value. So, the final intensity can be calculated as: \( I_f = 0.50 \times I_0 \)

As the intensity formulas are similar for both electric and magnetic fields, we can write the relation for the electric field as: \( \frac{I_f}{I_0} = \frac{E_f^2}{E_0^2} \) Similarly, we can write the relation for the magnetic field as: \( \frac{I_f}{I_0} = \frac{B_f^2}{B_0^2} \)

Replace the final intensity value from step 2 in the relations obtained in step 3: \( \frac{0.50 \times I_0}{I_0} = \frac{E_f^2}{E_0^2} \) \( \frac{0.50 \times I_0}{I_0} = \frac{B_f^2}{B_0^2} \) Now, cancel out the initial intensities: \( 0.50 = \frac{E_f^2}{E_0^2} \) \( 0.50 = \frac{B_f^2}{B_0^2} \) Next, we want to find the final electric and magnetic fields, i.e., \( E_f \) and \( B_f \). To do so, take the square root of both sides for the electric and magnetic field equations: \( E_f = E_0 \times \sqrt{0.50} \) \( B_f = B_0 \times \sqrt{0.50} \)

Finally, to express the reduction of the electric and magnetic fields as a percent, divide the final electric and magnetic fields by their initial values: \( \frac{E_f}{E_0} = \frac{E_0 \times \sqrt{0.50}}{E_0} \) \( \frac{B_f}{B_0} = \frac{B_0 \times \sqrt{0.50}}{B_0} \) Cancel out the initial electric and magnetic fields: \( \frac{E_f}{E_0} = \sqrt{0.50} \) \( \frac{B_f}{B_0} = \sqrt{0.50} \) Now, multiply by 100 to obtain the percentages: \( \frac{E_f}{E_0} \times 100 = 100 \times \sqrt{0.50} \) \( \frac{B_f}{B_0} \times 100 = 100 \times \sqrt{0.50} \) Approximate the calculations: \( \frac{E_f}{E_0} \times 100 \approx 70.71 \% \) \( \frac{B_f}{B_0} \times 100 \approx 70.71 \% \) Hence, the electric and magnetic fields are reduced to approximately 70.71% of their original values when the polarizing filter reduces the intensity of polarized light to 50.0% of its original value.