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Sagot :
The angle between the transmission axes of the polarizers if it is desired that one-tenth of the incident intensity be transmitted, ∝ = 63.435°
We'll assume that [tex]I_{0}[/tex] represents the incident light's intensity. Two polarizers are provided to us, but the first polarizer's angle is not disclosed. The incident light in situations like this needs to be unpolarized. This is due to the fact that, regardless of angle, the transmitted intensity via the polarizer is reduced by half for unpolarized light:
[tex]I_{1} =\frac{1}{2} I_{0}[/tex]
Then,
[tex]I_{2} =\frac{1}{10} I_{0}[/tex]
By using Malu's law,
[tex]I_{2} =I_{1} cos^{2} \alpha[/tex]
That is,
[tex]cos^{2} \alpha = \frac{I_{2} }{I_{1} }[/tex]
In terms of [tex]I_{0}[/tex], we get
[tex]cos^{2} \alpha =\frac{0.1I_{0} }{0.5I_{0} }[/tex]
[tex]cos^{2} \alpha = 0.2[/tex]
Now the angle is,
[tex]cos\alpha =\sqrt{0.2}[/tex]
cos ∝ = 0.44721
[tex]\alpha =cos^{-1} (0.44721)[/tex]
Then, ∝ = 63.435°
This angle is measured with respect to the first polarizer angle.
Learn more about the polarizers here: https://brainly.com/question/13008007
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