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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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6 J is the impulse caused by the change in velocity of 2 kg box from 2 m/s to 5 m/s.

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The magnitude of impulse is 6 J.

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As the mass of box is given as 2 kg and the velocity changes from 2 m/s to 5 m/s, then the impulse = 2 × (5-2) = 2 ×3 =6 J

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A 300 MHz electromagnetic wave in air (medium 1) is normally incident on the planar boundary of a lossless dielectric medium wit
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Answer:

Wavelength of the incident wave in air = 1 m

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Intrinsic impedance of media 2 = 125.68 ohms

Check the explanation section for a better understanding

Explanation:

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Speed of light in air, c =  3 * 10⁸ Hz

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Wavelength of the incident wave in medium 2

The refractive index of air in the lossless dielectric medium:

n = \sqrt{\epsilon_{r} } \\n = \sqrt{9 }\\n =3

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b) Intrinsic impedances of media 1 and media 2

The intrinsic impedance of media 1 is given as:

n_1 = \sqrt{\frac{\mu_0}{\epsilon_{0} } }

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n_1 = 377 \Omega

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n_2 = \sqrt{\frac{4\pi * 10^{-7} *1 }{8.84 * 10^{-12} *9 } }

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c) The reflection coefficient,r  and the transmission coefficient,t at the boundary.

Reflection coefficient, r = \frac{n - n_{0} }{n + n_{0} }

You didn't put the refractive index at the boundary in the question, you can substitute it into the formula above to find it.

r = \frac{3 - n_{0} }{3 + n_{0} }

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d) The amplitude of the incident electric field is E_{0} = 10 V/m

Maximum amplitudes in the total field is given by:

E = tE_{0} and E = r E_{0}

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