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Unpolarized light of intensity I=10 falls on two successive polarizer wheels with the angle between the polarizer wheels e-60°. What is the resultant intensity of the polarized light after leaving the 2nd polarizer? A) (0)/2 B) (10)/4 C) (0)/8 D) (o)/16

Respuesta :

Answer:

[tex]I=\frac{10}{4}[/tex]

Explanation:

A polarizer changes the orientation of the oscillations of a light wave.

I₀ = Intensity of unpolarized light = 10

θ = Angle given to the polarizer = 60°

Intensity of light

I = I₀cos²θ

⇒I = 10cos²60

[tex]\\\Rightarrow I=10\times \frac{1}{2}\times \frac{1}{2}\\\Rightarrow I=\frac{10}{4}[/tex]

So, the after passing through the second polarizer is [tex]\mathbf{\frac{10}{4}}[/tex]

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