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2 years ago

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Monochromatic light of wavelength, lambda, is traveling in air. The light then strikes a thin film having an index of refraction

, n1 that is coating a material having an index of refraction n2. If n1 is larger than n2, what minumim film thickness will result in minimum reflection of this light?A. lambda/(4*n2)B. lambda/n2C. lambda/4D. lambda(2*n1)E. lambdaF. lambda/(2*n2)G. lambda/n1H. lambda/(4n1)I. lambda/2

1 answer:

kumpel [21]2 years ago

3 0

Answer:

The correct option is H

Explanation:

From the question we are told that

The index of refraction of coating is $n_1$

The index of refraction of material is $n_2$

Generally the condition for constructive for a thin film interference is mathematically represented

$2 * t = [ m + \frac{1}{2}] \frac{\lambda}{n_1 }$

Here t represents the thickness

For minimum thickness m = 0

So

$2 * t =0 + \frac{1}{2}\frac{\lambda}{n_1 }$

=> $t =\frac{\lambda}{4n_1 }$

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A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

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3 years ago

In ice cap climates, the intense cold _____.

crimeas [40]

Answer:

A. is true

Explanation:

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2 years ago

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An object is allowed to fall freely near the surface of a planet. The object has an acceleration due to gravity of 24 m/s2. How

Alborosie

Answer:

12 m

Explanation:

The object is in uniformly accelerated motion, so the distance covered can be found using the following suvat equation:

$s=ut+\frac{1}{2}at^2$

where

s is the distance

u is the initial velocity

t is the time

a is the acceleration

For this problem,

$g=24 m/s^2$

and

u = 0, since we are considering the first second of motion

So, substituting t = 1 s, we find

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If there is an object in top of another object, why does the upper object exert a downward normal force?

rusak2 [61]

Answer:

This is because normal force is exerted perpendicularly to the point of contact between the upper and lower objects.

Explanation:

This is because the upper object is still subject to gravitational pull. Therefore, the amount of force it exerts on the lower object due to gravity will be equal to the normal force that acts in the negative direction of gravitational force. Additionally, normal force is evident because the upper object will not go into the lower object.

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planet a takes one year to go around the sun at a distance of one au. .planet b is three a u. from the sun. how many years does

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Answer:

3 years

Explanation:

Find the circumference of each orbit in AU.

2xπx1=6.283185307

2xπx3=18.84955592

Divide them.

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