Solving Problems About Radiation

2021-04-28 09:45:47
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Exercise 1.1 Intensity of Radiation

1. Assuming the radiation emitted by the cellular is in the form of spherical waves. What is the intensity of this radiation at the following distances:

a) 10 cm

Intensity = power / area

And area of a sphere = 4pr2 = 4 *p*0.12 = 0.12566m2 = ((300 x 10-3)/0.12566)W/m2 = 2.3874 W/m2

b) 100m

Intensity = power / area

And area of a sphere = 4pr2 = 4 *p*1002 = 125,663.7061 m2 = ((300 x 10-3)/ 125,663.7061)W/m2 = 2.3873 x 10-6 W/m2

c) 1km

1km = 1000m

Intensity = power / area

And area of a sphere = 4pr2 = 4 *p*10002 = 12,566,370.6144m2 = ((300 x 10-3)/12,566,370.614)W/m2 = 2.3873 x 10-8 W/m2

2. Give an expression for the time dependent electric field as a function of the distance r from the cellphone E = E(r,t) and use the expression to calculate the maximum electric field at the distances specified above.

E = Eo sin 2p/wavelength(x (velocity * time))

Velocity = distance / time

Therefore time and time cancel out and the remaining equation is as follows: E = Eo sin2p/wavelength(x - distance)

Therefore the electric field distances are as below: E = Eosin2p/wavelength(distance) = sin2p/400(0.1) = 0.0015707

E = Eosin2p/wavelength(distance) = sin2p/400(100) =1.5707

E = Eosin2p/wavelength(distance) = sin2p/400(1000) =15.707

3. Give a similar expression for the magnetic field produced by the cellular.

B = B(r,t)

Exercise 1.2 Exploration

 

1. To find the function for the co efficient we first get the fundamental period

Fundamental period To = 1

Omega wo = 2p/To =2p

Co =1/201 f(x).dx=1/2{00.5 1.dx + 0.51 0.dx} =0.5{0.5+0} = 0.25

However since k is not equal to 0

Ck = 1/ To = 1

Ck = 1 01 f(x) e -jk wo t.

= 1{00.5 1. e -jk wo t. + 0.51 0. e -jk wo t. } = 1{1/ -jk wo [e -jk wo t ]0 0.5 + 0/-jk wo [ e -jk wo t ] 0.5 1 } = 1{1/ -jk wo [e -jk wo 0.5 1 ] + 0/-jk wo [ e -jk wo t - e -jk wo 0.5 ]} = 1{1/ -jk wo [e -jk wo 0.5 1 ] + 0 ]}

Since wo = 2p

= 1{1/-jk2p [e -jk 2p 0.5 1] + 0. }

Since e -jk 2p 0.5 = cos k2p jsink2p = 1

=1{ 1/-jk2p [ 1-1]} = 1{1/-jk2p[0]} = 0

Therefore the co efficient for n = 1,2,3 are all zero as obtained from the equation above.

 

Exercise1.3 Cylindrical waves

Cylindrical waves are those produced by uniform line source since the wave front of such waves are cylindrical. The source of the wave is assumed to be an axis from which the waves are transmitted. If the source is an axis, then we would have a wave with cylindrical symmetry and all the wave particles at the same r distance from the source at the same time will be in the same phase.

There is inverse proportionality between the waves intensity and the distance from the source (r). The further away from the source, the lower the intensity and vice versa. Intensity is directly proportional to the square of amplitude. Therefore, amplitude is also inversely proportional to the square root of the distance from the source (r). Hence, amplitude is equal to Eo divided by square root of r where Eo is the amplitude at unit distance from the source.

The equation of cylindrical waves at radial distance r, from the source and at time t, would be:

Amplitude * cos (kr wt)

But amplitude = Eo/(r^0.5)

Therefore the equation would be Eo/(r^0.5)*cos (kr-wt) where r is distance from the source and w is angular frequency of the wave. This is a function of two variables, distance from source and time t.

E(r,t) = Eo/(r^0.5) * cos (kr-wt)

Exercise 1.4(Gaussian Optics)

b) The diameter of the minimum waist of the beam is given by:

2Wo = 2* ( wavelength*length/2pi)

Wavelength= 632nm

Length= 0.25m

Diameter = 2 * (0.000632mm * 250mm/2pi) = 0.5029mm

c) The diameter at the output coupler is given by

2 * Wo * ( 1 + (wavelength*(length from waist to the output coupler)/(pi*Wo^2))^2)^0.5

Length = 0.25/2 = 0.125 m = 125mm

2 * 0.5029mm * ( 1 + ((0.000632 mm * 125mm)/(pi*0.5029^2))^2)^0.5

= 1.0058mm *(1 + (0.079/0.7945)^2)^0.5 = 1.0108mm

d) distance of 2.75 m from the output coupler

length = 0.125+ 2.75 = 2.875m = 2875mm

beam diameter = 2 * 0.5029 * (1+ (( 0.000632 mm * 2875) /(pi*0.5029^2))^2)^0.5

= 1.0058 * ( 1 + 1.817/0.7945)^2)^0.5 = 2.5104mm

e) focal distance of 2.75 m

Yes theres some improvement due to the lens. The diameter is slightly smaller.

The diameter at the coupler is 1.0108 mm from above

To get the new diameter at 2.75m from coupler with lens:

= 4*wavelength/pi *focal distance/ diameter = 4 * 0.000632/pi * 2750/1.018 = 2.1892mm which is much smaller than 2.5104mm (from above)

f) The divergence half angle is given by

without lens

length = 2750mm + 250mm = 3000mm total distance

length from waist to 2.75m beyond coupler = 2750 + 125 mm = 2875mm

Wo = 0.5029mm

Diameter at 2.75m beyond the coupler is 2.5104 mm

Full angle divergence is (2.5104 0.5029)/ (3000-2875) = 0.01606

With lens

Wo = 0.5029mm

Diameter at the lens = 2.1892mm

Length = 2750 + 250= 3000

Length 2 = 2750 + 125 = 2875

Therefore = (2.1892-0.5029) / (3000-2875) =0.01349

 

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