Документ взят из кэша поисковой машины. Адрес оригинального документа : http://www.mso.anu.edu.au/~geoff/HEA/Q1-4.pdf Дата изменения: Wed Aug 5 03:39:25 2009 Дата индексирования: Tue Oct 2 00:21:46 2012 Кодировка:

High Energy Astrophysics Solutions to Questions 1-4
1. The relationship between energy and wavelength is hc 6.62 10 3.0 10 ­6 = ----- = ---------------------------------------------------- = 1.24 10 --- m ­ 19 1.60 10 -- m Also, to determine frequency and wavelength given energy, use 1.60 10 14 = h = -- = -------------------------- ------ = 2.41 10 ------ Hz ­ 34 eV eV h 6.62 10 hc hc ­6 = ----- = ----- = 1.24 10 ------ eV
­1 4 m = 1.24 10 ------ eV ­1 ­ 19 ­ 34 8 ­1

eV

Angstroms

Hence the following table (the primary quantity is given in bold):

Photon Typical microwave Typical far infrared

(Hz)

(Angstroms)

(metres) 0.20 100 10 = 10
­6

(eV) 6.2 10 1.1 10 0.62 2.48 12.4
­6 ­2

­4

Typical near infrared Typical optical Typical UV Soft X-ray Hard X-ray TeV -ray 2.41 10 2.41 10 2.41 10
17 19 26

2 10 5000 1000 12.4 0.124 1.24 10
­8

­6 ­7

5 10 10

­7 ­9

1.24 10 1.24 10 1.24 10

1 keV 100 keV 1TeV

­ 11 ­ 18


2.

R

c r

P

The flux density received at point P is: F =





I cos d =





I cos sin d d

= 2I

0



c

cos sin d = I sin2

c

R2 = I ----r2 That is, we recover the inverse square law.


3.
F


I


d = size of circular aperture

L

Circular aperture L sec d d cos Projected aperture d cos d

First calculate the solid angle of rays hitting the P back of the camera. Viewed from the point P the aperture appears as an ellipse with major and minor axes d and d cos . The area of the aperture is therefore A -- d cos d = -- d 2 cos 4 4 Hence the (small) solid angle subtended by the aperture at P is 2 -- d cos d2 4 = ---------------------- = -- ----- cos3 4 L2 L sec 2



Next, the flux density incident on the back of the camera at P is

F =





4 -- cos ----- 4 d2 4 I cos d I cos = I -- 2 cos = I ----------------4L f2

where the focal ratio f = L d . 4. (a) The mean intensity at a particular point, a radius r from the centre of the sphere is: 11 J = ----- I d = -4 2


0



c

1 1 R2 I sin d = -- I 1 ­ cos c = -- I 1 ­ 1 ­ ----- 2 2 r2

1/2

(b) Flux through the surface of the sphere is:


n I


F =

0 0

2





I cos sin d d = I



Hence, the total energy per unit time through the surface of the sphere (i.e. the Luminosity) is: L = F 4 R 2 = 4 2 I R
2

Therefore, L I = -------------42R2 (c) The energy density per unit frequency is 4 4 1 R2 -u = ----- J = ----- -- I 1 ­ 1 ­ ----- c c2 r2 L R2 = --------------- 1 ­ 1 ­ ----- 2 cR 2 r2 L --------------4 cR 2 Hence, the total (i.e. frequency integrated) energy density is Lu --------------4 cR 2 (d) This expression can also be derived as follows. The total energy passing through a sphere of radius r is u c 4r and this is equal to the total luminosity, L . Hence L u c 4 r 2 = L u = -------------4 cr 2 (e) For L = 3. 83 10
26 2 1/2 1/2

W and r = 1.50 10 u = 4.5 10
­6

11

m = 2.8 10
13

Jm

­3

eV m

­3