### Video Transcript

If a polarizing filter reduces the intensity of polarized light to 50.0 percent of its original value, what is the ratio of the amplitude of the electric and magnetic fields in the light incident from and emergent from the filter?

Letβs begin by drawing a sketch of the light going into and leaving the filter. In this sketch of a ray of light moving through an optical filter, the perpendicular electric and magnetic fields are represented by capital πΈ and capital π΅, respectively. If we let πΌ sub π be the initial intensity of the ray of light and πΌ sub π be its final intensity after passing through the filter, then weβre told that a 50 percent of πΌ sub π is equal to πΌ sub π. We can now recall a fact connecting light intensity with the amplitude of both the electric and magnetic fields that make up that ray of light. Intensity πΌ is proportional to the square of the magnitude of the electric field amplitude, which is also proportional to the square of the magnitude of the magnetic field amplitude.

Focusing just on the electric field for simplicity, if we call πΈ sub π the electric field amplitude before the field reaches the filter and πΈ sub π the electric field amplitude after the filter, then we can write that πΌ sub π divided by πΌ sub π is proportional to πΈ sub π squared divided by πΈ sub π squared. We can factor out the square in the numerator and denominator of the right side. And we can rewrite πΌ sub π divided by πΌ sub π as 2.00 because of the given relationship between those two values. If we take the square root of both sides, the squared term and square root cancel out on the right. And we find that πΈ sub π over πΈ sub π equals 0.707, the square root of 2.00. This means that the ratio of the initial to the final electric field amplitude is equal to one to 0.707. And remember, this ratio is true also for the magnetic fields.