浙江大学大学物理甲下chapter

A1 Amin
A2 , I max A1 A2 , Imin
Φ 2 n2r2 n1r1 2
m,
n2r2
n1r1
(2m 1)
2
,
m 0,1,2...Imax m 0,1,2...Imin
(1) Double-slit interference
nor1 nor2 d sin m
m 1,2,3,...
The angle θ is the angle between the incident x-ray and the plane.
m 0, 1, 2, 3,... (maximum)
d sin (m 1)
2 m 0, 1, 2, 3,... (minimum)
d sin d tan d y m
D
D
ym m d m 0,1,2,3,... (maxima)
y
ym1
ym
D
d
(2) Interference from thin film
2d
n22
n12sin
2i
2
,
m
m 1,2,3......max ima
(2m
1)
2
m 0,1,2......min ima
If i =0, 2nd
2
( 2d n ,
n
2
in the film)
Example 1: Two slips interference, S1S2=0.7mm, D=100cm, λ=500nm. Now the single slip is shifted and SS1-SS2= λ/2, an optical medium plates with thickness of l and index of
Exercise Lesson
Wave optics
1. Interference Phase difference and optical path difference
Φ
2
n2r2
2
2
n1r1
1
( Let 1 2 )
2
(n2r2
n1r1 )
2m ,
(2m
Amax
1) ,
refraction of 1.5 placed at the back of one slits. The point o is 4th minimum fringe. Find (1) l, (2) the distance the central
maximum to the point o.
Solution:
(1) SS2 S2O l nl (SS1 S1O)
(2m 1) , m 3
SS2
2 SS1 l(n 1)
Fra Baidu bibliotek
7
2
D
l(n 1) 7 l 4μm
22
(2) SS2 r2 l nl (SS1 r1) 0
l(n 1) 2 r2 r1 0
r2
r1
[l(n
f
x0
2f
1
2f
a
width of the central maximum
x f ,
a
width of other fringes
(2) Grating diffraction
d sin m m 0,1,2,... Principle maxima
(3) Resolving power
Airy spot:
n=1.4 n=1.5
Solution: for maximum , 2nd m,
at the edge of film , d 0, is bright.
for minimum, 2nd (2m 1) , m 0,1, 2
2
If the 1st minimum correspond to m=0, the center point is
2d
2. Diffraction (1) Single-slit diffraction
0
Bright fringe at the center
asin
2m
2
m
m 1,2,3...min ima
(2m
1)
2
m 1,2,3...max ima
a sin a tan a a xm m
m=20.
2nd (40 1) , d 4100nm
2
If the 1st minimum correspond to m=1, the center point is
m=21. 2nd (2m 1) , m 1, 2
2
Example 4 (E41-40): 2d(n 1) 60, n 1 60 1.00030
R
min
1.22
d
,
R 1/R resolution ability
Grating:
D m d cos
R mN
(4) missing order
a sin m1 d sin m2
d a
m2 m1
m2
d a
m1
is missing
(5) Bragg’s law
2d sin m
1)
2
]
7
2
r2
r1
d D
x
x
D d
(r2
r1)
D d
7
2
2.5mm
r1
x
r2
Example 2: There is an oil film (n2=1.20) on flat grass plate (n1=1.52). A light ray (λ=600nm) illuminates normally and is observed from above by reflected light. Find the thickness of the film at 5th bright fringe from the edge of oil film.
Solution:
for maximum , 2n2d m
m 0,1, 2
At the edge of film, d=0 (m=0), is bright.
Thus the 5th bright fringe is m=4. d m 1.010 6 m
2n2
Example 3: As shown, λ=560nm, center point is dark, there are 20 dark rings in the outer region. Find the thickness of the film at center.