1主极大(中央明纹中心) I rsin(using/n wsin6/元 =0处,Ip=l0(=Imx) aa e 2次极大(次级明纹中心 d/0→tga=a;图解法→ =aidd, y tga a << 用半波带 法可解释 2兀 元 半波带法较粗略 振幅矢量法精确十 c=-3.47x,-2.46兀,-1.43, +1.43兀,+2.46兀,+3.47兀, →asin6=-347,-2.46λ,-1.43, +1.43元,+2.46,+3.47元 1p=0.008310,0.01650,0.0472l0, 0.04720,0.0165L0,0.008310, 3极小(暗纹中心:asi=土(k=1,2,…)处,Ip=0;
o -2 - 2 y y1 = tg y2 = · · · · 2 0 ] sin / sin( sin / ) [ a a I I 1.主极大(中央明纹中心): P = 0 , ( max ) 0 I I I = 处 P = = 3.极小(暗纹中心): sin = ( = 1,2, ) = 0; P a k k 处，I = -3.47π, -2.46π, -1.43π, +1.43π, +2.46π, +3.47π, asin = −3.47 , − 2.46, −1.43, + 1.43, + 2.46, + 3.47, 2.次极大(次级明纹中心): , 0 tg ; sin = = → = d a dI 图解法 0.0472I0 , 0.0472I0 0.0165I , 0 , 0.0165I0 ,0.0083I0 0.0083I , 0 = , P I I0 , 半波带法较粗略 振幅矢量法精确 I次<<I0 用半波带 法可解释
「例1单缝的夫琅禾费衍射中=4=30°对应的是暗纹还是明纹? 解:(用半波带法)asin的/(2/2)=4 对应的是暗纹 「例2在单缝的夫琅禾费衍射中=0.6mm,f=40cm,x 14mm处为一明纹求:(1)入射光波长;(2)该明纹 的级次;(3)对该明纹而言缝平面被分为几个半被带? 解:(2k+1)·(λ/2)=a(x/)=21×10m k=1,=14×10-6m;(x) k=3,元=0.6×10m;7个半波带k=2,4=0.84×106m;(x) k=4,元=0467×106m;-9个半波带k=5,=0382×10m;(× [例3]如图示雷达射束波长λ=30mm 求:雷达监视范围内公路的长度L d=15 解:将波束看成单缝衍射的0级明纹「902m a·sin,=A,→s、30mm=0.15,→≈8.63?→ 0.20m c=15°+61=23.63,B=15°-61=6.37: →,L=d(ctgB-ctga)=15(ct6370-cg23.63°)≈100m 作业2.425
[例1]单缝的夫琅禾费衍射中, , a = 4 对应的是暗纹还是明纹? o = 30 解: (用半波带法) asin /( / 2) = 4 对应的是暗纹. 解: k a x f m 6 (2 1) ( / 2) ( / ) 2.1 10− + = = 1, 1.4 10 ;( ) 6 = = − k m 2, 0.84 10 ;( ) 6 = = − k = 3, = 0.610−6m; k m 4, 0.467 10 ; 6 k m − = = 5, 0.382 10 ;( ) 6 = = − k m 7个半波带 9个半波带 d=15 m a=0.20m β L θ1 150 [例3]如图示雷达射束波长λ=30mm. 求:雷达监视范围内公路的长度L. a sin 1 = , = 15(ctg6.37°− ctg23.63°) 100m = = = 0 15, 0 20m 30mm sin . a . 1 1 8.63? 15° 23 63, 1 = + = . 15° 6 37; 1 = − = . ,L = d(ctg − ctg) 解: 将波束看成单缝衍射的0级明纹 [例2]在单缝的夫琅禾费衍射中,a = 0.6mm, f = 40cm, x = 1.4mm处为一明纹.求: (1)入射光波长; (2)该明纹 的级次; (3)对该明纹而言,缝平面被分为几个半被带? 作业 2.4 2.5