Problem is from Physics for scientists and Engineers withModern Physics by Gianc

Question

Problem is from Physics for scientists and Engineers withModern Physics by Giancoli ch.35 # 16 Design a double-slit apparatus so that the central diffractionpeak contains precisely seventeen fringes. Assume the firstdiffraction minimum occurs at (a) an interference minimum (b) aninterference maximum. I don't get how the 9th fringe gets canceled out by the firstfringe minimum and how to really start this problem ingeneral Problem is from Physics for scientists and Engineers withModern Physics by Giancoli ch.35 # 16 Design a double-slit apparatus so that the central diffractionpeak contains precisely seventeen fringes. Assume the firstdiffraction minimum occurs at (a) an interference minimum (b) aninterference maximum. I don't get how the 9th fringe gets canceled out by the firstfringe minimum and how to really start this problem ingeneral

Explanation / Answer

Given :     The central diffraction peak contains 17fringes. So there will be 8 fringes on each side of the center.Thus the 9th interference maxima of the double slit mustcoincide with the first minimum of the difference pattern resultingin the missing order of 9th fringe. The condition forinterference maxima d sin = m                                                                  where m = 0,±1,±2.....                                      let m = 9                           d sin = 9    So,                        sin = 9/d --- ( 1 )                The condition for minima in a single- slit diffractionpattern d sin = m                                                                  where m = 0,±1,±2.....                                      let m = 1                           d sin = 1    So,                        sin = /d ---- ( 2 ) From eqn ( 1 ) & ( 2 )                 d = 9D I hope it helps you                                                                  where m = 0,±1,±2.....                                      let m = 9                           d sin = 9    So,                        sin = 9/d --- ( 1 )                The condition for minima in a single- slit diffractionpattern d sin = m                                                                  where m = 0,±1,±2.....                                      let m = 1                           d sin = 1    So,                        sin = /d ---- ( 2 ) From eqn ( 1 ) & ( 2 )                 d = 9D I hope it helps you                                                                  where m = 0,±1,±2.....                                      let m = 1                           d sin = 1    So,                        sin = /d ---- ( 2 ) From eqn ( 1 ) & ( 2 )                 d = 9D I hope it helps you

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.