浙江大学大学物理甲下chapter

(n2r2
n1r1)

2m , Amax (2m 1) ,
A1 Amin
A2 , I max A1 A2 , I min
Φ 2 n2r2 n1r1 2


m,


n2 r2
n1r1

(2m 1)

2
,
m 0,1,2...Imax m 0,1,2...Imin
(1) Double-slit interference
nor1nor2 dsinm
m0,1,2,3,..(.maximum)
dsin (m1)
2 m0, 1, 2, 3,...(minimum)
dsindtandym
D
ymmdD m0,1,2,3,.(..maxima)
0.05 nm is barely resolved. Meanwhile there are no other
principle maxima.
Solution: dsinm, d 2 240n0m
sin30
R mN, N 6000

m
When λ=400nm,
2
no
ne
d
Quarter-wave plate and circular polarization*
Quick quiz
Two ideal polarizing sheets are stacked so that none of the incident unpolarized light is transmitted. A third polarizing sheet is slipped between the first two sheets at an angle 45ºto the bottom sheet. The fraction of light transmitted through the entire stack is
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)S2SS2Oln l(S1SS1O)

(2m1) , m3


SS2
2 SS1 l(n1)

7 2

D
l(n1) 7 l 4μm
22
(2)S2Sr2ln l(S1Sr1)0
l(n1)2r2r10
r2

r1

[l(n
k=-1, 0, 1, 2, 4, 5, these 6 fringes are observed.
Example 6: Assume that the limits of the visible spectrum are arbitrarily chosen as 400 and 760nm. Calculate the number of rulings per mm of a grating that will spread the first-order spectrum through an angle range of 20.0°.
Solution:
fom r ax,im 2 un2m dm
m0,1,2
At the edge of film, d=0 (m=0), is bright.
Thus the 5th bright fringe is m=4. dm1.0106m
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.
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.
si n m d , 2 d 7 0 3 d 4 0,0 s0 i 2 ,7 n 0 s 0i 3 ,4 n 0
No matter what the value of d, the above formula come into existence.
3. Polarization
sin 1 10 sin 20 0.356 cos 1 19 10 cos 20 1 tan 1 0.356 19 36 d 1 1.192 10 3 mm
sin 19 36 N 1 839 /mm
d
Example 7: E43-22
Exercise Lesson
Wave optics
1. Interference Phase difference and optical path difference
Φ

2
n2 r2


2


2
n1r1

1

( Let 1 2 )

2
(1) Law of Malus: I= Im cos2θ
Polarizing sheet
(2) Brewster’s law
tap n n n 1 2
(3) Double refraction:
n 1 w 1 ,hte ap n n2
no and ne
no ne d,
2
Example 8: To design a grating, when it is illuminated at
normal incidence by white light, the light with λ=600nm is
viewed in second order at 30°, and the doublet of Δλ =
m=20.
2nd (4 01), d41n0m 0
2
If the 1st minimum correspond to m=1, the center point is
m=21.

2nd (2m1) ,
m12,
2
Example 4 (E41-40):
2d(n1)60, n1601.00030
(5) Bragg’s law
2dsin m
m 1,2,3,...
The angle θ is the angle between the incident x-ray and the plane.
Example 5: (E43-12) Assume that light is incident on a grating at an angle φ as shown. Show that the condition for a diffraction maximum is d(sin φ ±sinθ) =mλ.
yym1ymdD
(2) Interference from thin film
2d
n22
n12sin2i


2
,
m
m1,2,3.....m. axima
(2m1)

2
m0,1,2.....m. inima
If i =0, 2nd
(2dn, in
Solution: (a) condition (1) is satisfied,
dsi3n 0 2 2, ds2 i 3 2 n 0 24 n0 m 0
(b) condition (3) is satisfied, d/a=3, a=800nm.
(c) dsinθ=mλ2, let θ￫90°, m=4. All orders are 0, ±1, ± 2. The ± 3 is missing order, the ± 4 is not observed actually.
f
x02f12f a
w oifd tc h th e enm traax l im

xf ,
d tow ho f ithfreirn ges
a
(2) Grating diffraction
d si n mm 0 , 1 , 2 ,.P.ri.nciple maxima
Solution:
1s9i1n 1s0i1 n 2 ()0 1 (0 s1 c i2 n o 0 c so 1 s2 s i ) n 0
1 s9 1 i 1 n s0 1 i c n 2 o 1 0 c s0 1 o s2 i s n 0
If d=2μm, λ=590nm, φ=30º, d=3a, how many fringes can be observed?
d(si nsin )m , 30 , , m 5.08
2
d3a, m 3ismissing.
d (s s in i ) n m , 3 ,0 , m 1 .7 2
dsin m , 240 si0 3n0 m 40 0m 3
d3,a80 n0 m a
Example 9: E43-11
Solution: If the second order spectra overlaps the third order, it is because the 700nm second order line is at a larger angle than the 400nm third order line.
2d
2. Diffraction (1) Single-slit diffraction

0
Brighftringaet thceenter
asin 2m2 m
m1,2,3...minima
(2m1)2
m1,2,3..m . axima
asinatanaaxmm
2
nle 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
n=1.4 n=1.5
Solution: fom r axi,m 2nudm ,
at tehdegoefil,m d0,isbrig . ht
fo m r i,n 2 n i m d (2 m 1 u ),m m 0 ,1 2 , 2
If the 1st minimum correspond to m=0, the center point is
(3) Resolving power
Airy spot:
R min1.22d, R1/R resolutiaobnility
Grating:
D dcmos
R mN
(4) missing order
asinm1 dsinm2
dam m12 m2dam1 ismissing