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Lunna [17]
2 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]2 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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For a 50 kg person receives an absorbed dose of gamma radiation of 20 millirads,  the total energy absorbed is mathematically given as

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<h3>What is the total energy absorbed?</h3>

Generally, the equation for the total energy absorbed  is mathematically given as

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E=50*20*19^{-3}

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Coulomb's law:

  Force = (<span>8.99×10⁹ N m² / C²<span>) · (charge₁) · (charge₂) / distance²

            = (</span></span><span>8.99×10⁹ N m² / C²<span>) (1 x 10⁻⁶ C) (1 x 10⁻⁶ C)  / (1.0 m)²

            = (8.99×10⁹ x 1×10⁻¹² / 1.0)  N

            =      8.99×10⁻³  N

            =        0.00899 N repelling.

Notice that there's a lot of information in the question that you don't need.
It's only there to distract you, confuse you, and see whether you know
what to ignore.

-- '4.0 kg masses';  don't need it. 
   Mass has no effect on the electric force between them.

-- 'frictionless table';  don't need it.
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A proton is shot perpendicularly at an infinite plane of charge. The charge density of the plane is +7.65×10^−4 C/m^2. If the pr
Alexeev081 [22]

Answer:

The velocity is 2.94\times10^{6}\ m/s.

Explanation:

Given that,

Charge density of the plane \sigma=7.65\times10^{−4}\ C/m^2

Distance = 1.05 mm

We need to calculate the electric field due to plane of charge

Using formula of electric field

E=\dfrac{\sigma}{2\epsilon}

Put the value into the formula

E=\dfrac{7.65\times10^{−4}}{2\times8.85\times10^{-12}}

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We need to calculate potential difference

Using formula of potential difference

V=E\times r

Put the value into the formula

V=4.322\times10^{7}\times1.05\times10^{-3}

V=4.5381\times10^{4}\ Volt

We need to calculate the work requires to be done to reach the surface of the plane

Using formula of work done

W=qV

Put the value into the formula

W = 1.6\times10^{-19}\times4.5381\times10^{4}

W=7.26096\times10^{-15}\ J

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Using work energy theorem

W=\dfrac{1}{2}mv^2

v^2=\dfrac{2W}{m}

v=\sqrt{\dfrac{2\times7.26096\times10^{-15}}{1.67\times10^{-27}}}

v=2.94\times10^{6}\ m/s

Hence, The velocity is 2.94\times10^{6}\ m/s.

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