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spayn [35]
1 year ago
10

What is the maximum resultant intensity in terms of the intensity of the initial unpolarized incoming light io?

Physics
1 answer:
Dennis_Churaev [7]1 year ago
6 0

The unpolarized light becomes less intense. When unpolarized light is run through a polarizer, its intensity is diminished by a factor of ½.

<h3>Unpolarized light:</h3>

Unpolarized light is diminished by a factor of ½ when it travels through a polarizer. Cos 2θ averages out across all angles to be ½ in size. I transmitted = I₀ and ½ I₀ at all angles The unpolarized light's intensity decreases to zero.

The light-producing atoms on the heated filament's surface behave independently of one another. Each of these emissions can be roughly represented as a brief "wave train" lasting between approximately 109 and 108 seconds. These wave trains, each with its unique polarization direction, are superimposed to form the electromagnetic wave that is coming from the filament.

Unpolarized light is produced by all common light sources, including the Sun, incandescent and fluorescent lighting, and flames.

Learn more about unpolarized light here:

brainly.com/question/15562548

#SPJ4

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Eld a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1
zheka24 [161]

Answer:

v_{f}=1721.1m/s

Explanation:

Given data

q = 3.1 µC

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As we know that:

dK=-dU\\(1/2)mv_{f}^{2}-(1/2)mv_{i}^{2}=-(\frac{kq^{2} }{r_{f}}-\frac{kq^{2} }{r_{i}} )\\    (1/2)*(47*10^{-6}kg )v_{f}^{2}-(1/2)*(47*10^{-6}kg)(0)=-[(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{2.5*10^{-3}}-(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{0.83*10^{-3}} )]\\2.35*10^{-5} v_{f}^{2}=69.61\\v_{f}=\sqrt{\frac{69.61}{2.35*10^{-5}} }\\v_{f}=1721.1m/s

5 0
3 years ago
Calculate the electric field at the center of a square
pantera1 [17]

Answer:

E_y=1175510.2\ N.C^{-1}

The Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

Explanation:

Given:

  • side of a square, a=52.5\ cm
  • charge on one corner of the square, q_1=+45\times 10^{-6}\ C
  • charge on the remaining 3 corners of the square,q_2=q_3=q_4=-27\times 10^{-6}\ C

<u>Distance of the center from each corners</u>=\frac{1}{2} \times diagonals

diagonal=\sqrt{52.5^2+52.5^2}

diagonal=74.25\cm=0.7425\ m

∴Distance of center from corners, b=0.3712\ m

Now, electric field due to charges is given as:

E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}

<u>For charge q_1 we have the field lines emerging out of the charge since it is positively charged:</u>

E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}

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<u>Force by each of the charges at the remaining corners:</u>

E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}

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<u> Now, net electric field in the vertical direction:</u>

E_y=E_1-E_4

E_y=1175510.2\ N.C^{-1}

<u>Now, net electric field in the horizontal direction:</u>

E_y=E_2-E_3

E_y=0\ N.C^{-1}

So the Magnitude of electric field is in the upward direction as shown directly towards the charge q_1.

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