Re: [題目] 雙狹縫干涉

看板Physics作者 (rachel5566)時間13年前 (2011/03/09 23:07), 編輯推噓1(100)
留言1則, 1人參與, 最新討論串5/6 (看更多)
※ 引述《hbkhhhdx2006 (大T)》之銘言: : [領域] 雙狹縫干涉 : [來源] 課本習題 : [題目] Monochromatic light is directed on a double slit. A light meter is : placed to the right of the slits. When slit S2 is closed, the : light intensity at the location of the meter is I1. When slit S2 is : closed the light intensity is I2. : (a) What is the light intensity "I" when both slits are open if : x1-x2=λ? : (b) What is "I" if x1-x2=(1/2)λ? : Assume that the size of the slits is smaller than the wavelengh so : that the slits can be considered to be point sources. : 附圖:http://ppt.cc/DYgQ : [瓶頸] 這題我完全不知道怎麼解 : 原文書上沒有提到這種東西以前也沒學過 : 答案是 (a) I1-I2+2(I1I2)^(1/2) (b) I1+I2-2(I1I2^(1/2) 因為光強度與振幅平方成正比,故可令 I1 = k(A1)^2 和 I2 = k(A2)^2 當 x1 - x2 = λ,也就是完全建設性干涉,其振幅 A = |A1 + A2| 總強度 I = k(A1 + A2)^2 = k(A1)^2 + k(A2)^2 + 2k(A1)(A2) = I1 + I2 + 2[(I1)(I2)]^(1/2) 當 x1 - x2 = (1/2)λ,也就是完全破壞性干涉,其振幅 A = |A1 - A2| 總強度 I = k(A1 - A2)^2 = k(A1)^2 + k(A2)^2 - 2k(A1)(A2) = I1 + I2 - 2[(I1)(I2)]^(1/2) -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 140.112.211.87

03/09 23:47, , 1F
謝謝!! 答案還打錯真不好意思
03/09 23:47, 1F
文章代碼(AID): #1DTvUZHL (Physics)
文章代碼(AID): #1DTvUZHL (Physics)