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?
Let’s 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 𝐸 and capital 𝐵, respectively. If we let 𝐼 sub 𝑖 be the initial intensity of the ray of light and 𝐼 sub 𝑓 be its final intensity after passing through the filter, then we’re told that a 50 percent of 𝐼 sub 𝑖 is equal to 𝐼 sub 𝑓. 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 𝐼 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 𝐸 sub 𝑖 the electric field amplitude before the field reaches the filter and 𝐸 sub 𝑓 the electric field amplitude after the filter, then we can write that 𝐼 sub 𝑖 divided by 𝐼 sub 𝑓 is proportional to 𝐸 sub 𝑖 squared divided by 𝐸 sub 𝑓 squared. We can factor out the square in the numerator and denominator of the right side. And we can rewrite 𝐼 sub 𝑖 divided by 𝐼 sub 𝑓 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 𝐸 sub 𝑖 over 𝐸 sub 𝑓 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.