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Lunna [17]
3 years ago
15

Unpolarizedlight of intensity I_0 is incident on three polarizingfilters. The axis of the first is vertical, that of the secondi

s 60.0 ^\circ from vertical, andthat of the third is horizontal.
What light intensityemerges from the third filter? (I0)
Physics
1 answer:
Marina86 [1]3 years ago
5 0

To solve this problem it is necessary to apply the concepts related to the law of Malus which describe the intensity of light passing through a polarizer. Mathematically this law can be described as:

I = I_0 cos^2\theta

Where,

I_0 = Indicates the intensity of the light before passing through the polarizer

I = Resulting intensity

\theta= Indicates the angle between the axis of the analyzer and the polarization axis of the incident light

From the law of Malus when the light passes at a vertical angle through the first polarizer its intensity is reduced by half therefore

I_1= \frac{I_0}{2}

In the case of the second polarizer the angle is directly 60 degrees therefore

I_2 = I_1 cos^2\theta

I_2 = (\frac{I_0}{2} ) cos^2(60)

I_2 = 0.125I_0

In the case of the third polarizer, the angle is reflected on the perpendicular, therefore, its angle of index would be

\theta_3 = 90-60 = 30

Then,

I_3 = I_2 cos^2\theta_3

I_3 = 0.125I_0 cos^2 (30)

I_3 = 0.09375I_0

Then the intensity at the end of the polarized lenses will be equivalent to 0.09375 of the initial intensity.

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