Question

(4) Use the preliminary observations to answer these questions; Compared to no polarizer or analyzer in the optical path, by what percent does the light intensity decrease when (a) The polarizer is introduced into the optical path? (b) The both polarizer and analyzer are introduced into the optical path?

Answer 1

a) I = I₀/2, b) I = I₀/2 cos² θ

Step-by-step explanation:

To answer these questions, let's analyze a little the way of working of a polarized

* When a non-polarized light hits a polarizer, the electric field that is not in the direction of the polarizer is absorbed, so the transmitted light is

i = I₀ / 2

and is polarized in the direction of the polarizer

* when a polarized light reaches the analyzer it must comply with Malus's law

I = I₁ cos² θ

where the angle is between the polarized light and the analyzer.

With this, let's answer the questions

a) When a polarizer is placed in the non-polarized light path, half of it is absorbed and only the light that has polarization in the direction of the polarizer is transmitted with an intensity of

I = I₀/2

b) when a polarizer and an analyzer are fitted, the intensity of the light transmitted by the analyzer is

I = I₀/2 cos² θ

where the final value depends on the angle between the polarizer and the analyzer.

Let's look at two extreme cases

θ = 0 I = Io / 2

θ = 90º I = 0