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Anettt [7]
3 years ago
5

Unpolarized light of intensity i0 is incident on three polarizing filters. the axis of the first is vertical, that of the second

is 43° from vertical, and that of the third is horizontal. what light intensity emerges from the third filter?
Physics
1 answer:
Eva8 [605]3 years ago
7 0
To determine the intensity, the key idea is to work through the system filter-by-filter and applying either the one-half rule for unpolarized light or the cosine-squared rule for an already polarized light.

When the light reach the first filter, it is unpolarized, so the intensity is :
            I1 = 1/2 I0                     (One-half rule)

Because the light reaching the second filter is polarized, given 43° angle difference, the intensity is : 
            I2 = I0 cos^2 (angle difference)
                = I0 cos^2 43°
                = 0.535 I0

Lastly, when the light reached the third filter, it is polarized and has angle diference equals to the angle between the second and last filter 47°. Using the same rule, we get :
           I3 = I2 cos^2 47°
               = (0.535 I0) cos^2 47°
               = (0.535 I0) (<span>0.465)
               = 0.249 I0

So, the intensity which emerges is 0.249 I0.</span>
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The temperature of the resistor varies based on the variation in the temperature. The equation that describes the relationship between the two of them is:

R = R0[1+ alpha(T-T0)]  where:

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Substitute in the equation to get R as follows:

R = 0.0200 [1+( ₋ 4.8 x 10^{-4}) (0-25.8)] = 0.02020 ohm

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