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Butoxors [25]
3 years ago
6

Polarized light passes through a polarizer. If the electric vector of the polarized light is horizontal what, in terms of the in

itial intensity I0, is the intensity of the light that passes through a polarizer if the polarizer is tilted 22.5° from the horizontal?
Physics
1 answer:
rewona [7]3 years ago
3 0

Answer: I0*0.853

Explanation:

Ok, the Malus's law says that:

If you have light polarized along a given line with an intensity I0, and it passes through a polaroid which axis of polarization forms an angle θ with respect to the polarization of the light, then the intensity of the resulting beam is:

I(θ) = I0*cos^2(θ)

For example, if the axis of the polaroid is exactly the same as the axis of polarization of the light beam that will impact it, then we have θ = 0°, and the equation above says that the intensity of the beam will not change.

In this particular case, we have that the intensity of the light is I0, and the angle is θ = 22.5°

Then:

I(22.5°) = I0*cos^2(22.5°) = I0*0.853

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3 years ago
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A current of 02kA is traveling through a circularly looped wire. The wire makes 35 turns around an induced magnetic field of 8 9
PIT_PIT [208]

Answer:

The radius of the loop is 4.94\times10^{5}\ m

Explanation:

Given that,

Current = 0.2kA = 200 A

Number of turns = 35

Magnetic field = 8.9 nT

We need to calculate the radius of one loop

Using formula of magnetic field

B=\dfrac{N\mu_{0}I}{2r}

r=\dfrac{N\mu_{0}I}{2B}

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B = magnetic field

r = radius

Put the value into the formula

r=\dfrac{35\times4\pi\times10^{-7}\times200}{2\times8.9\times10^{-9}}

r =494183.11\ m

r=4.94\times10^{5}\ m

Hence, The radius of the loop is 4.94\times10^{5}\ m

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3 years ago
How do you do this? I know that the Density equation is D=m/v but these are in kilograms and Liters...so do I convert it to gram
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4 years ago
A circular conducting loop with a radius of 0.50 m and a small gap filled with a 10.0-Ω resistor is oriented in the xy-plane. If
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Answer:

Current, I = 0.153 A

Explanation:

Given that,

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To find,

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\phi_f\ and\ \phi_i are final flux and the initial flux respectively.

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The magnitude of current can be calculated using the Ohm's law as :

I=\dfrac{\epsilon}{R}

I=\dfrac{1.53}{10}

I = 0.153 A

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4 years ago
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