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pav-90 [236]
2 years ago
12

Light of intensity I0 and polarized horizontally passes through three polarizes. The first and third polarizing axes are horizon

tal, but the second one is oriented 20.0� to the horizontal. In terms of I0, what is the intensity of the light that passes through the set of polarizers?
A) 0.442 I0
B) 0.180 I0
C) 0.780 I0
D) 0.883 I0
Physics
1 answer:
Alex17521 [72]2 years ago
3 0

Answer:

Option C.

Explanation:

Suppose that we have light polarized in some given direction with an intensity I0, and it passes through a polarizer that has an angle θ with respect to the polarization of the light, the intensity that comes out of the polarizer will be:

I(θ) = I0*cos^2(θ)

Ok, we know that the light is polarized horizontally and comes with an intensity I0

The first polarizer axis is horizontal, then the intensity after this polarizer is:

then θ = 0°

I(0°) = I0*cos^2(0°) = I0

The intensity does not change. The axis of polarization does not change.

The second polarizer is oriented at 20° from the horizontal, then the intensity that comes out of this polarizer is:

I(20°) =  I0*cos^2(20°) = I0*0.88

And the axis of polarization of the light that comes out is now 20° from the horizontal

Now the light passes through the last polarizer, which has an axis oriented horizontally, so the final intensity of the light will be:

note that here the initial polarization is  I0*0.88

and the angle between the axis is 20° again.

Then the final intensity is:

I(20°) =  I0*0.88*cos^2(20°) = I0*0.78

Then the correct option is C.

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The second ball traveled a greater distance when compared to the first ball because the second ball spent more time in motion.

The given parameters;

  • time of fall of the first ball, t = 1 s
  • time of fall of the second ball, t = 3 s

The distance traveled by each ball is calculated using the second equation of motion as shown below.

The distance traveled by the first ball is calculated as follows;

h = u_0t + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 1^2)\\\\h = 4.9 \ m

The distance traveled by the second ball is calculated as follows;

h = \frac{1}{2} gt^2\\\\h = (0.5\times 9.8\times 3^2)\\\\h = 44.1\ m

Thus, the second ball traveled a greater distance because it spent more time in motion.

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3 0
1 year ago
The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad
kenny6666 [7]

Answer:

a)  v = 2,9992 10⁸ m / s , b)  Eo = 375 V / m ,  B = 1.25 10⁻⁶ T,

c)     λ = 3,157 10⁻⁷ m,   f = 9.50 10¹⁴ Hz ,  T = 1.05 10⁻¹⁵ s , UV

Explanation:

In this problem they give us the equation of the traveling wave

        E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]

a) what the wave velocity

all waves must meet

        v = λ f

In this case, because of an electromagnetic wave, the speed must be the speed of light.

        k = 2π / λ

        λ = 2π / k

        λ = 2π / 1.99 10⁷

        λ = 3,157 10⁻⁷ m

        w = 2π f

        f = w / 2 π

        f = 5.97 10¹⁵ / 2π

        f = 9.50 10¹⁴ Hz

the wave speed is

        v = 3,157 10⁻⁷   9.50 10¹⁴

        v = 2,9992 10⁸ m / s

b) The electric field is

           Eo = 375 V / m

to find the magnetic field we use

           E / B = c

           B = E / c

            B = 375 / 2,9992 10⁸

            B = 1.25 10⁻⁶ T

c) The period is

           T = 1 / f

            T = 1 / 9.50 10¹⁴

            T = 1.05 10⁻¹⁵ s

the wavelength value is

          λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm

this wavelength corresponds to the ultraviolet

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