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3 years ago

A light beam has speed c in vacuum and speed v in a certain plastic. The index of refraction n of this plastic is

Physics

1 answer:

Mnenie [13.5K]3 years ago

3 0

Answer:

$n = \frac{c}{v}$

Explanation:

Refractive Index: It is a measure to find how fast the light travels through a medium. It is ration of the speed of light in vacuum to speed of light in the medium. Speed of light is not constant and varies depending on the density of the medium.

In vacuum the speed of light is 300000 km/s and is denoted by c. When the light beam enters any medium the speed will decrease. Here it is given that the speed in plastic is v. Thus the refractive index(n) is given as:

$n = \frac{c}{v}$

It is a dimensionless no.

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are the properties of a specific element the same as the properties of a molecule made up of that element ? why or why not

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3 years ago

Red laser light from a He-Ne laser (λ = 632.8 nm) creates a second-order fringe at 53.2° after passing through the grating. What

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Explanation:

It is given that,

Wavelength of red laser light, $\lambda=632.8\ nm=632.8\times 10^{-9}\ m$

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For diffraction grating,

$d\ sin\theta=n\lambda$

$d=\dfrac{n\lambda}{sin\theta}$ , n = 2

$d=\dfrac{2\times 632.8\times 10^{-9}}{sin(53.2)}$

$d=1.58\times 10^{-6}\ m$

The wavelength λ of light that creates a first-order fringe at 22 is given by :

$\lambda=d\ sin\theta$

$\lambda=1.58\times 10^{-6}\ sin(22)$

$\lambda=5.91\times 10^{-7}\ m$

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Hence, this is the required solution.

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4 years ago

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

$f=12GHz=12\cdot 10^9 Hz$ is the frequency

$c=3.0\cdot 10^8 m/s$ is the speed of light

So the wavelength of the beam is

$\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m$

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

$y=\frac{m\lambda D}{a}$

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

$y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m$

and so, the diameter is

$d=2y = 750 m$

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

$r=\frac{750 m}{2}=375 m$

So the area is

$A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2$

And since the power is

$P=100 kW = 1\cdot 10^5 W$

The average intensity is

$I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2$

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3 years ago

Weak magnetic fields can be measured at the surface of the brain. Although the currents causing these fields are quite complicat

STALIN [3.7K]

To develop this problem it is necessary to apply the concepts related to a magnetic field in spheres.

By definition we know that the magnetic field in a sphere can be described as

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Our values are given as

$D=2a = 16cm \rightarrow$ diameter of the sphere then,

$a = 0.08m$

Thus z = a

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$B = \frac{\mu_0I}{2(2^{3/2})a}$

$B = \frac{\mu_0 I}{2^{5/2}a}$

Re-arrange to find I,

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Therefore the current at the pole of this sphere is $1.08*10^{-6}A$

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3 years ago

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Answer:

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3 years ago