Question

A plane polarized light is incident normally on the tourmaline plate. its \(\mathrm{E}^{\rightarrow}\) vectors make an angle of \(45^{\circ}\) with the optical axis of the plate. find the percentage difference between initial and final maximum values of \(\mathrm{E}^{\rightarrow}\) vectors. (A) \(19 \%\) (B) \(92 \%\) (C) \(50 \%\) (D) \(29 \%\)

Short answer

The percentage difference between the initial and final maximum values of the electric field vector is closest to (C) $50\%$.

Step-by-step solution

Recall Malus's law

Malus's law states that the intensity of the transmitted light through a polarizer depends on the angle between the electric field vector and the optical axis of the polarizer. Mathematically, it can be expressed as: \[I = I_0 \cdot \cos^2(\theta)\] where: - \(I\) = intensity of the transmitted light, - \(I_0\) = intensity of the incident light, and - \(\theta\) = angle between the electric field vector and the optical axis of the polarizer

Determine the initial and final maximum values of electric field vectors

Since we are given that the initial angle between the electric field vector and the optical axis of the plate is 45 degrees, we can use Malus's law to calculate the initial value of the electric field vector. The electric field vector is proportional to the square root of the intensity: \(E = \sqrt{I}\) The initial and final maximum values of the electric field vectors can be found by substituting the given angle into Malus's law: \[E_i = \sqrt{I_0 \cdot \cos^2(45^{\circ})}\] \[E_f = \sqrt{I_0}\]

Calculate the percentage difference between the initial and final electric field vectors

To calculate the percentage difference between the initial and final values of the electric field vectors, we can use the following formula: \[\frac{E_f - E_i}{E_i} \times 100\%\] Substitute the values of \(E_i\) and \(E_f\) from the previous step: \[\frac{\sqrt{I_0} - \sqrt{I_0 \cdot \cos^2(45^{\circ})}}{\sqrt{I_0 \cdot \cos^2(45^{\circ})}} \times 100\%\] Now, simplify the expression: \[\frac{\sqrt{I_0} - \sqrt{I_0 \cdot \frac{1}{2}}}{\sqrt{I_0 \cdot \frac{1}{2}}} \times 100\%\] \[\frac{\sqrt{I_0} - \sqrt{\frac{I_0}{2}}}{\sqrt{\frac{I_0}{2}}} \times 100\%\] Finally, calculate the numerical value: \[\frac{\sqrt{2} - 1}{1} \times 100\%\] \[\approx 41.42\%\] So, the percentage difference between the initial and final maximum values of the electric field vector is around \(41.42\%\). However, the options given are: (A) \(19 \%\) (B) \(92 \%\) (C) \(50 \%\) (D) \(29 \%\) The closest option to the calculated value is (C) \(50\%\). Although the exact value is not among the given answers, we can select the closest option as the final answer.