Respuesta :
Answer:
percentage of unpolarized light intensity,
[tex]\frac{I"'}{I} = 8.52%[/tex]
Given:
Polarization angle for the 2nd polarizer, [tex]\theta = 50.9^{\circ}[/tex]
Solution:
If I be the intensity of unpolarized light, then the intensity of light passing through the first polarizer, I':
[tex]I' = \frac{I}{2}[/tex]
Now, the intensity of light through polarizer 2, I":
[tex]I" = I'cos^{2}\theta = \frac{I}{2}cos^{2}\theta[/tex]
Also Polarizer 1 and 3 are perpendicular to each other:
Therefore, [tex]\theta' = 90^{\circ} - 50.9^{\circ} = 49.1^{\circ}[/tex]
Intensity of light passing through polarizer 3:
[tex]I"' = I" cos^{2}\theta' = \frac{I}{2}cos^{2}\theta cos^{2}\theta' [/tex]
[tex]I"' = \frac{I}{2}cos^{2}50.9^{\circ}\times cos^{2}49.1^{\circ}[/tex]
I"' = 0.0852 I
Now, percentage of unpolarized light intensity:
[tex]\frac{I"'}{I}\times 100 = 0.0852\times 100 = 8.52%[/tex]
