物质在绝对温度T下
1. The focal length of the components is 7 mm , -15 mm, and the spare between them
are 1.5 mm. (a) What is the focal length of the lens system? (b) If the distance of the
object is 30 mm, calculate the distance of the image.
f=7mmf=-15mm
30mm
1.5mm
Ans:
m1=[1 0
-1/7 1];
m2=[1 1
0 1];
m3=[1 0
1/15 1];
mt=m3*m2*m1;
f=-1/mt(2,1)
mt
a=[1 1/30]
b=mt*a'
s=-b(1)/b(2)
Results:
f = 11.6667
mt =
0.8571 1.0000
-0.0857 1.0667
a =
1.0000 0.0333
b =
0.8905
-0.0502
s =
17.7532
2. (a) Please explain Boltzmann Distribution
,,N,Cexp,EKTn.
(b)Draw two lines to show Boltzmann probability Distribution at
temperature T and T (T < T). 1212
(c) Discuss two case: temperature T =0 and T=, 12
Ans:
E
2E
E1
E0
N
物質在絕對溫度T下,在不同的能階的原子數目
物質在絕對溫度T下,其原子數在不同的能階的數目N:圖6-3-2:可用以下數學式表示:
,,N,Cexp,EKTn
其中,K為波茲曼常數
C為常數
En第n個能階之能量值
從上式得知,在平常時候,高能階中的原子數目或居量:Population:應較低能階之居量為小。
(b) 在高溫時候,高能階中的原子數目或居量:Population:應較低溫時候之居量為多。
E
T1
2E T2
(T < T).12
1E
0EN
(c) 在非常高溫:T=,:的時候,分佈曲線呈現直豎的情形,高能階中的原子數2
目或居量:Population:與低能階中的原子之居量幾乎一樣多。在非常低溫:T =0 :1的時候,分佈曲線呈現躺帄的情形,高能階中的原子數目幾乎沒有,所有原子處於最低能階中。
E
2E
T2
1E
T1
0EN
A two-slit Young’s interference experiment is arranged as illustrated in
figure 1; the wavelength of incident light is ,=500nm. When a transparent material is put behind one of the slits, the zero order fringe
moves to the position previously occupied by the 4th order bright fringe.
The index of refraction of the transparent material is n=1.2. Calculate the
thickness of the transparent material.
n=1.2
Figure 1
Ans:
Intensity maxima occur when the optical path difference is ,=m,.
Thus
,,=,,m
When the transparent material is inserted as shown is Fig. 1, the change of optical path difference is given by
,,=t(n-1)
where t is the thickness of the transparent material. As the interference pattern shifts by 4 fringes.
,m=4
Hence t(n-1)=4,, giving
4λ t,,10,mn-1
A Michelson interferometer is adjusted to give a fringe pattern of concentric circles when illuminated by an extended source of light of ,=500nm.
(a) How far must the movable arm be displaced for 1000 fringes to
emerge from the center of the bullseye?
(b) If the center is bright, calculate the angular radius of the first dark ring
in terms of the path length difference between the two arms and the
wavelength ,.
Ans:
Concentric circles, called fringes of equal inclination, are obtained when the two reflecting mirrors are exactly mutually perpendicular. The optical path-length difference , between the two arms is given by
,,2ndcos,,2n(l,l)cos, 12
where n is the index of refraction of the medium (for air, n=1), d=l-l is 12the optical path-length difference (OPD) between the arms of lengths l 1and l, , is the angle of incidence of light at the mirrors. 2
(1) For 1000 fringes to emerge from the center of the bullseye, the OPD ,
must undergo a change of 1000,. Thus 2d=1000, or ,
d=500=0.25mm, i.e., the movable arm is displaced 0.25mm. ,
(2) If the center is bright, we have 2d=m ,where m is an integer. ,
For the first dark ring, we have
12dcos,,(m-), 2
After subtracting yields
,2d(1-cos),, 2
2,,cos1-For small ,, , hence ,2
,, rad. ,2d
For d=0.25mm, we obtain ,=0.032 rad=1.8:
1. 請舉實例說明稜鏡三大用途.
(a)色散:分光光譜儀;(b)倒像:照像機與雙筒望遠鏡倒像之用;(c)改變光的
進行方向:光學儀器合光分光之用
2. 求像距與像高.
(a)一凸透鏡焦距10 inch,物位於第一焦點左邊40 inch,物高5 inch.
(b) 一凸透鏡焦距10 inch,物位於第一焦點右邊2 inch,物高5 inch.
答案(a) 像位於第二焦點右邊2.5 inch, 像高 -1.25 inch.
(b) 像位於第二焦點左邊50 inch, 像高25 inch.