[0:00]Hello students, how are you? Welcome to my YouTube channel Pankaj Physics Guruji. Students, aapko is channel par na keval graduation physics ke topics ki sabse asaan explanation dekhne ko milti hai, balki aapko saath-saath yeh bhi bataya jaata hai ki aapko exam mein kya aur kaise likh ke aana hai taaki aap sirf video dekh ke hi apne notes bana sakein aur exams mein bahut acche marks laa sakein.
[0:20]Toh students, aaj is channel ko subscribe kijiye aur paaiye notification meri nayi video ka sabse pehle. Isi category mein aage badhte hue aaj hum baat karenge BSc third sem, Wave and Optics One, Unit three ke ek aur important topic ke baare mein jiska naam hai Diffraction at a Straight Edge.
[0:39]Toh students, diffraction at a straight edge ko samajhne se pehle aapko diffraction ke baare mein pata hona chahiye ki diffraction kya hota hai.
[0:46]Uske kya parameters rehte hain, kya conditions hoti hain diffraction phenomena occur hone ki? Iske upar main already video upload kar chuka hoon, Diffraction of Light ke naam se. Yeh video agar aap dekh lenge toh aapko yeh topic jo hai bahut asaan lagega samajhne mein.
[1:01]Kyunki jab tak aapko diffraction nahi pata hoga toh aap iske concept ko samajh nahi paayenge ki straight edge se diffraction kaise hota hai. Toh meri aapse prarthna hai ki pehle aap yeh video dekh ke is topic ko karne aaiye.
[1:13]Aur is topic ka, is video ka link main aapko niche description mein de dunga, wahan jaake aap isko dekh sakte hain. Toh students, diffraction hota hai bending of light around the corners of an obstacle or an aperture or a slit.
[1:25]Diffraction jo hai, woh light ka mud jaana kisi bhi slit ya kisi aperture ke corners se that is called diffraction. Toh diffraction at a straight edge ko consider karne ke liye humein sabse pehle chahiye hoga ek source of light jo humein monochromatic light dega wavelength jiski hogi lambda.
[1:43]Aur us monochromatic light ko pass karwaya jaayega ek rectangular narrow slit se jo ki humne yahan par S li hai. Aur yeh humari straight edge hai jahan se humne diffraction karwana hai. Yahan par iska nukila sira hoga yahan par, yahan se diffraction hoga, is edges se diffraction hoga.
[2:00]Iska example aap straight edge ka le sakte hain razor blade ki form mein. Yeh humari screen hai jahan par humein diffraction pattern obtain karna hai. Toh yeh humara freshnel class of diffraction chal raha hai jismein humara source aur slit ke beech ka jo distance rehta hai woh finite rehta hai.
[2:17]Aur humara jo screen aur slit ya phir aperture ya kisi bhi obstacle ka jo distance rehta hai screen se woh bhi finite rehta hai. Toh yeh dono distance jo hain woh hum measure kar sakte hain. Yeh wala distance bhi humein pata hoga, yeh humara distance humne A maan liya. Aur straight edge se jo humara screen ka distance hoga woh bhi finite hoga, usko hum B maan lenge.
[2:35]Toh students, jab monochromatic light wavelength Lambda ke is narrow slit mein se niklegi toh woh cylindrical wave front emit karegi aur is cylindrical wave front humne XY maan liya jo is straight edge ke upar incident hoga.
[2:48]Aur means straight edge ko humne ek, is cylindrical wave front XY se illuminate karwaya hai. Toh students hoga kya? Yahan par aakar is light ka jo hai woh diffraction ho jaayega.
[2:59]Aur agar geometrical optics ki baat karein jo humein yeh kehti hai ki light hamesha straight line mein chalti hai, toh edge ke niche toh light block ho jaayegi. Toh screen ke upar jo light hai aapko upar wala portion dikhayi dega.
[3:10]Means agar hum screen ke mid point par ek point le lein P jo ki bilkul ek hi seedh mein hoga, toh P ke upar jo hai aapko poori tarah se illumination dikhayi deni chahiye geometrical optics ke hisab se aur P ke niche jo hai aapko darkness dikhayi deni chahiye.
[3:25]But actual practice mein aisa nahi hota. Actual practice mein aapko P ke upar uniform illumination nahi dikhayi deti jabki wahan par aapko dikhayi dete hain alternate dark and bright bands of unequal width and varying intensity.
[3:39]Alag-alag width ke, alag-alag intensity ke aapko yahan par dark and bright bands dikhayi denge. Aur jaise-jaise hum P se upar jaate jaayenge, jaise-jaise hum P se upward direction mein jaate jaayenge, waise-waise jo woh bands hain dark and bright bands unki intensity bhi kam hoti jaayegi.
[3:52]Aur woh close bhi hote jaayenge aur unko alag-alag kar paana mushkil ho jaayega aur kuch distance ke baad, ek particular distance pe pahunchne ke baad woh dark and bright bands jo hain woh disappear ho jaayenge aur uske baad aapko complete illumination dikhayi degi.
[4:09]P point ke niche aapko complete darkness dikhayi deni chahiye thi jabki aisa nahi hoga kyunki light jo hai woh yahan se diffract hoke thoda niche tak bhi pahunchegi, geometrical shadow region mein bhi pahunchegi aur thodi si aapko illumination dikhayi degi uske baad aapko darkness dikhayi degi.
[4:24]Toh in cheezon ka explanation hum denge. Aur hum dekhenge ki screen ke kisi bhi point par aapko agar dark band milega ya bright band milega uski conditions kya hongi, yeh sab kuch dekhenge hum aaj ke topic ke andar.
[4:36]Toh students, agar aapko aata hai ki discuss diffraction at a straight edge, toh aap starting mein kya likh ke aayenge theory mein? Aaiye woh dekh lete hain.
[4:44]Consider a straight edge AB which is illuminated by monochromatic light of wavelength lambda from a narrow slit source S. Let small A be the distance of source S from AB and B be the distance of screen from AB. XY is the cylindrical wavefront incident on the edge. According to laws of geometrical optics, we expect illumination above P and a complete darkness below P on screen. But in practice, it is observed that, first, just above P, there are alternate dark and bright bands of unequal width and varying intensity. These bands becomes closer and less distinct as we further move upwards away from P and ultimately they disappear giving an uniform illumination.
[7:20]Toh students, last section mein humne dekha ki P point se upar jo hai aapko alternate dark and bright band dikhayi denge. Ab P se niche kya hoga? Woh dekhte hain. Below point P: In geometrical shadow region, the intensity falls off continuously and rapidly until complete darkness is reached a short distance below P. Explanation of diffraction bands.
[8:42]Toh students, agar aap P point se upar nazar aane wale alternate dark and bright bands ko explain karna hai toh aapko use karna padega concept of Freshnel half period zones.
[8:53]Dekhiye students, Freshnel half period zones kya hote hain, woh Freshnel ne kyun construct kiye diffraction phenomena ko explain karne ke liye aur kaise construct kiye jaate hain Freshnel half period zones. Uske upar main already detailed video bana chuka hoon jiska aap logon ne behad hi usko bahut zyada usko sneh diya us video ko. Un videos ke link main aapko niche description mein de dunga Freshnel half period zone part one, part two karke banaya tha.
[9:17]Woh video jo hai bahut hi important topic hai is unit ka aur wahan jaake aap description mein jaake wahan se iske links dekh sakte hain aur tab jaake aapko complete samjh aa sakta hai ki Freshnel half period zones kya hote hain.
[9:28]Dekhiye students, agar aapne alternate dark and bright bands ko explain karna hai P point se upar, toh is screen XY ko aapko bahut saare half period zones mein divide karna padega. Pehla half period zone AM1, dusra half period zone M1 M2, teesra half period zone M2 M3, aise hi aage M3 M4 hoga. Aur jab aap half period zones mein divide karte hain, toh is tarah divide karte hain agar ki A se le ke P tak ka distance agar B hai, toh M1 se le ke P tak ka jo distance hoga, in dono distances mein half of wavelength ka difference hoga.
[10:04]Yeh humne wahan padha tha Freshnel half period zones wale mein abhi aapko yahan par yaad rakhna hai yeh cheez ki jo AP agar B hai toh M1 P hoga B+lambda/2. Aise hi M2 P mein lambda/2 aur add kar dijiye toh yeh ban jaayega B+2lambda/2. Aise hi M3 P distance jo hai woh pichle wale distance se lambda/2 aur zyada hoga toh yeh ho jaayega B+3lambda/2 kyunki 2lambda/2 mein agar lambda/2 add karenge toh 3lambda/2 milega, toh aise hi successive jo hain zones jo hain woh lambda/2 distance se separate honge.
[10:30]Isliye inko half period zones bola jaata hai. Kyunki successive zone se nikalne wali waves ke beech mein jo path difference hota hai, woh lambda/2 ka hota hai. Agar path difference lambda/2 ka hai toh phase difference jo hai woh pi nikal ke aayega.
[10:53]Means do successive half period zone se nikalne wali waves jo hoti hain woh out of phase hoti hain aur means unke beech mein phase difference jo hai woh pi ka hota hai. Is cheez ka bhi hum aage use karenge jab hum intensity find karenge screen ke upar kisi bhi point ke upar screen ke upar.
[11:08]Toh humne kaafi saare half period zones mein divide kar diya. Pehla half period zone AM1 ko hum first half period zone bolenge. M1 M2 ko hum bolenge second half period zone, M2 M3 ko bolenge third half period zone. Half period zone ko main HPZ se likhunga short form mein. Aur such is tarah aapne construct karne hain half period zones such that AP jo humara B hai toh M1 P jo hoga woh lambda/2 zyada hoga AP se B+lambda/2 ho jaayega. M2 P isse lambda/2 zyada hoga toh B+2lambda/2 ho jaayega. M3 P B+3lambda/2 ho jaayega and so on.
[11:38]Toh students, ab hum find karenge pehle hum P point par intensity find karenge ki kitni intensity hogi. Phir P point se upar ek point lenge P prime wahan dekhenge kitni intensity hogi, woh case bhi discuss karenge.
[11:47]Phir hum baat karenge geometrical shadow region ki aur saare jo cheezein hain one by one discuss karne ja rahe hain. Toh pehle hum baat karenge intensity at point P ki. Dekhiye students, P point ke liye jo humare paas is wave front XY ka sirf upar wala portion jo hai AX, upper half jo hai iska, AX sirf wahi exposed hai.
[12:00]Kyunki AY wave front jo hai woh is edge ke dwara block kar diya jaayega. Toh pehle aap yeh likhenge for point P on screen, pole of the wave front is at A. Agar P point par intensity nikalni hai toh XY ka jo mid point hoga woh A hoga. Means wave front ka jo pole hoga usko hum A point par maan ke chalenge. For point P, only upper half AX of the wave front XY is exposed, whereas lower half AY of the wave front XY has been blocked by the edge AB.
[12:23]P point ke liye AX portion exposed hoga jabki AY jo hai woh block ho jaayega is straight edge ke dwara. Toh hum kaise nikalenge ab aage intensity point P ke upar woh dekhenge hum ab next section ke andar.
[12:45]Toh students, last section mein hum find kar rahe the intensity at point P ke upar. Usi ko aage continue karte hain usmein aap aur kya add karke aayenge? Aaiye dekhte hain. Let A1, A2, A3 up to An be the amplitudes of at point P due to waves coming from first, second, third and up to nth half period zones of exposed part AX of wave front. The resultant amplitude at point P will be A = A1 - A2 + A3 - A4 and so on up to An.
[13:06]Toh students, pehle half period zone se jo wave nikal ke aa rahi hai, uska amplitude hai maan lijiye A1 aur doosre half period zone M1 M2 se jo wave nikal ke aa rahi hai, uska amplitude humne maan liya A2. Aur aise hi jo humara nth half period zone hoga jo humne yahan par construct kiya hai AX portion ke andar.
[13:23]Usmein se jo wave nikal ke aayegi uska jo amplitude rahega woh An hoga. Toh humare paas total resultant amplitude kitna rahega P point ke upar? Wahi humne find karna hai. The resultant amplitude at point P will be. Toh resultant amplitude point P par hoga in nth half period zones ke karan.
[13:37]Usko humne capital A maan liya aur woh un sabhi A1, A2, A3 se leke An ka sum hoga. Means capital A jo hoga woh A1+A2+A3+A4+A5 up to An hoga. But students, aapko ek cheez maine pehle bhi batayi thi, yeh half period zones ki baat kar rahe hain.
[13:54]Aur half period zones mein se jo waves nikalti hain, kisi bhi do successive half period zone mein se jo waves nikalti hain, unke beech mein path difference lambda/2 ka hota hai. Means phase difference jo hai woh pi ka hota hai. Means dono out of phase hoti hain.
[14:13]Aur means agar pehli wave ka amplitude A1 hai toh doosre half period zone se nikli hui wave ka jo amplitude hoga woh minus A2 lena padega. Humein sign jo hain woh opposite lene hain. Phir hum third wale ka positive lenge, phir fourth wale ka negative, aise aapko plus, minus, plus, minus jo hain woh leke chalne hain.
[14:30]Kyunki successive half period zone se nikalne wali waves ke beech mein phase difference jo hai woh pi ka hoga, out of phase hongi woh toh sign ka difference aapko lena padega. Toh capital A kya hoga A1, phir doosre wale ka out of phase kar dena hai minus A2, phir agle wale ka phir sign change kar dena hai plus A3, minus A4 up to An. Yeh wahan par humne detail mein discuss kiya tha Freshnel half period zone wale video mein.
[14:50]Wahan aap dekh sakte hain isko woh poori details ke andar. Toh students, humne kya karna hai? A1 ko do parts mein divide kar dena hai. A1 can be written as A1/2 + A1/2. Kyunki jab aap in dono ko sum karenge toh aapko A1 hi milega. Jitne bhi odd amplitudes hain A1, A3, A5, jitne bhi odd number wale amplitudes honge unko aapne do parts mein divide kar dena hai.
[15:09]Minus A2 aise hi rahega. A1 wale amplitude mein change karna hai aapne. A3 ko hum divide kar sakte hain A3/2 mein plus A3/2 mein. Minus A4 aise hi rahega. Aise hi iske aage A5 hoga toh A5/2, agle mein bhi A5/2 aayega toh humne ek A5/2 likh diya. Up to so on. Toh students, ab hum isko brackets mein put karte hain, rearrange kar dete hain isko.
[15:26]Yahan se leke yahan tak ek bracket mein daal dete hain aur yahan se leke yahan tak ek bracket mein daal dete hain. Dekhiye students, jaise-jaise hum upar jaate jaayenge P point se, in wave se nikalne wale amplitude jo hain woh kam hote jaayenge. Successive half period zone se jo amplitudes honge woh kam hote jaayenge, unka contribution jo hai woh P point par kam hota jaayega. A1 zyada hoga A2 se, aur A2 jo hai woh A3 se zyada hoga. A3 jo hai A4 se zyada hoga. Means decrease hote jaayenge.
[15:56]Matlab inka jo order rahega woh aisa rahega, A1 greater than A2, greater than A3, greater than A4 and so on. Means yeh ek prakar ka arithmetic progression form kar lenge. Agar teen terms humare paas A1, A2, A3, yeh AP mein hain. Toh humare paas A2 kya hota hai? A2 ko hum aise likh sakte hain A1 + A3 / 2.
[16:17]Jo humare middle term hoti hai, woh first and third term ka mean hoti hai. Yeh property hoti hai arithmetic progression mein. Agar teen numbers humare paas arithmetic progression mein hain, toh middle number can be written as the mean of first and last number. Means, is cheez ko aap use karenge yahan pe.
[16:32]Toh hum A2 ko kaise likh sakte hain? Pehle aap line likhenge, The amplitudes A1, A2, A3 decreases regularly. Jo amplitudes hain half period zones ke woh kam hote jaate hain lagataar. Isliye yeh ek prakar ki AP form karte hain, arithmetic progression form karte hain aur arithmetic progression ki property hoti hai ki jo humare middle term hoga, woh pehle usse pehle wale aur uske baad wale number ka jo hota hai woh amplitude, mean hota hai.
[16:54]Toh A2 kya hoga? A2 se pehle hoga A1, A2 se baad mein hoga A3 / 2. Isliye A4 kya hoga? A4 se pehle kya hota hai? A3 aur A4 ke baad mein kya hota hai? A5 / 2. Humne even numbers walo ko is form mein likhna hai. Toh A2 ko hum aise likh sakte hain, A4 ko aise likh sakte hain. Jab aap A2 ki value A1 + A3 / 2 yahan put karenge, A1 + A3 / 2 toh minus sign ki wajah se A1 / 2 minus ka ho jaayega aur A3 / 2 bhi minus ka ho jaayega.
[17:20]Toh yeh poori term jo hai woh zero ho jaayegi. Aap yahan put karke dekhiye A2 ki value yeh wali toh aapko yahan par zero milega. Isi prakar se jab aap A4 ko yahan par put karenge, A4 ki jo yeh wali hai value hai A3 + A5 / 2 woh yahan put karenge toh yeh bracket zero ho jaayegi. Agli bracket jo hogi humare paas usmein aapka A6 aa raha hoga.
[17:34]Agar A6 ki jagah aap uski value put karenge toh saari brackets jo hain woh aise zero ho jaayengi. Toh sirf aapke paas sirf ek term bachegi, woh hogi A1 / 2. Toh means resultant amplitude jo hoga P point par is portion AX ke karan.
[17:46]Is portion mein present jitne bhi half period zones honge unke karan, woh A1/2 hoga. Toh A jo hoga amplitude resultant amplitude P point par woh approximately A1/2 ke barabar hoga. Kyunki intensity jo hai woh amplitude ke square ke barabar hoti hai, toh intensity jo hai woh P point par iska square ho jaayegi. Means A1 square by 4 intensity aapko P point par mil jaayegi.
[18:09]Toh humne P point par intensity find kar li hai. Next section mein hum find karenge ek aise point ke upar intensity jo ki P se thoda sa upar hoga. Jaise ki P dash hum maan sakte hain jo ki P se upar hai. Yahan par aapko zyada distance lag raha hai par actual practice mein P dash jo hai woh P ke bilkul close hota hai. Means ab hum find karenge intensity at point P dash.
[18:27]Toh students, chaliye find karte hain intensity kisi aur point par P dash point par jo ki P se upar hai, just above hai aur humne P aur P dash ka distance jo hai woh X assume kar liya hai. Toh second point mein aap kya likhenge? Intensity at a point just above P.
[18:43]Toh students, agar aap P dash point par intensity nikalna chahte hain toh sabse pehle aapko S aur P dash ko join karna padega. Yeh jo line hai SP dash is wave front XY ko C point par intercept karegi. Iska matlab yeh hai ki P dash point ke liye C humara naya pole hoga wave front XY ke upar.
[19:01]P point ke liye jo pole tha woh A point tha aur P dash point ke liye pole jo hoga humara woh C hoga. Jaise humne P point par intensity find karne ke liye is wave front ko bahut saare half period zones mein divide kar diya tha, first, second, third half period zones. Usi tarah se P dash point par intensity find karne ke liye XY wave front ko bahut saare half period zones mein divide kar diya jaayega.
[19:25]Ab jo intensity hai P dash point par woh maximum hogi ya minimum hogi? Is baat pe depend karega ki aapka jo portion hai AC iske andar kitne half period zones included hain. Agar ek hai toh intensity kuch aur hogi, agar AC portion ke andar half period, number of half period zones ka number two hai toh P dash par intensity kuch aur hogi. Is prakar se depend karega P dash point par ki AC portion ke andar half period zones ki jo jo number hai woh even sankhya mein hai ya odd number mein hai.
[19:56]Is baat pe depend karega aur hum dekhenge ki jab even ya phir odd number of half period zones honge AC portion ke andar toh P dash point par intensity jo hai woh kya hogi aur woh kaise hum nikalenge? Yahi humne is point mein dekhna hai. Toh student, chaliye students, dekhte hain ki aap is point mein kya likh ke aayenge.
[20:11]Now consider a point P dash on screen just above P. Let P be at a distance X from P such that only one half period zone of wavefront CA is exposed. So, resultant amplitude A1 at P dash will be A1 = A1 / 2 + A1 = 3A1 / 2. I1 = (3A1 / 2)^2 = 9A1^2 / 4 - I maxima.
[23:36]Now if point P dash is moved through some distance (upward) w.r.t. P such that 2 half period zones of wavefront CA are exposed, then resultant amplitude at P will be A2 = A1 / 2 + A1 - A2 = (A1 / 2) - A2. I2 = (A1 / 2 - A2)^2 - I minima.
[25:48]Toh students, last section mein humne dekha ki agar ek half period zone hoga toh intensity P dash point par hogi 3A1/2 ka whole square. Agar do half period zones honge toh intensity hogi P dash point par 3A1/2 - A2 ka whole square. Similarly, I3 = (3A1 / 2 - A2 + A3)^2 - II maxima.
[26:02]I4 = (3A1 / 2 - A2 + A3 - A4)^2 - II minima. And so on. Thus, intensity at a point above P will be, (1) maximum if number of HPZ in CA is odd. (2) minimum if number of HPZ in CA is even. Toh students, intensity find kar chuke hain hum.
[27:33]Uske baad hum find karenge positions of maxima and minima aur yeh kaam karenge hum ab next section ke andar. Toh students, next heading jo aap banayenge woh banayenge Positions of maxima and minima. Pichle section mein humne find kiya ki first maxima, second maxima and so on, first minima, second minima and so on unki intensities kya hongi kisi bhi point ke upar jo ki P se above hoga screen ke upar.
[27:57]Toh students, ab humein find karni hai maximum and minima ki positions. Dekhiye students, kisi bhi point P dash ke upar maximum banega ya minimum banega yeh is baat pe depend karega ki A point se and C point se nikalne wali waves ke beech mein kitna path difference hai.
[28:10]Aur hum jaante hain diffraction phenomena ke andar jab path difference jo hai light waves ke beech mein woh odd multiple of lambda/2 ka hota hai toh aapko maxima dekhne ko milta hai. Toh usmein kya likhenge aap? To obtain maxima at P dash we have path diff = (2n + 1)lambda/2, n = 0,1,2.
[29:27]Toh agar humne AP dash find karna hai toh humein iske liye use karni padegi right angle triangle AP P dash. Toh humara right angle triangle AP P dash mein yeh humara base hai, yeh humara perpendicular hai. Humein find karna hai AP dash means hypotenuse find karna hai. Aur jo hypotenuse hota hai woh hota hai square root of base ka square that is B square plus perpendicular ka square that is X square.
[29:58]Aur square root ko hum aise bhi likh sakte hain power 1/2 ki form mein bhi likh sakte hain toh yeh humari third second equation ho gayi. Humne AP dash find kar liya hai, bahut asaan tha find karna. AP P dash triangle mein agar aap Pythagoras theorem lagayenge toh aap badi asani se AP dash find kar sakte hain kyunki aapko AP bhi pata hai, PP dash bhi pata hai.
[30:19]Toh AP dash humara nikal gaya. Toh CP dash hum nikalne ke liye humein kya karna padega? CP dash ko hum kaise likh sakte hain? SP dash - SC. SP dash - SC. Toh SC dekh lete hain kya hai? Dekhiye students, jo A aur C points hain woh bilkul close hain ek doosre ke. Aapko figure mein yahan thoda distance nazar aa raha hoga, but actual practice mein C point is very close to A. Toh aise distance aur jo SA distance hoga usko hum barabar maan sakte hain. SC approximately equal to SA. Toh humein SC ka SA ka distance pata hai that is A.
[30:42]The distance between the edge and the source jo ki small A hai toh SC ko bhi hum small A ke barabar maan sakte hain. Toh SC ko hum A maan lenge aur SA humara hai A. SC is approximately equal to SA. Toh SP dash humein find karna hai. Isko hum third equation de dete hain. SP dash jo hai woh is right angle triangle se niklega SPP dash.
[31:10]Toh humara is triangle ke andar hypotenuse hai woh hum SP dash, jo hypotenuse hota hai woh hota hai base ka square plus perpendicular ka square ki power 1/2. Toh humara base kitna hai? Base hai humara SP. Toh hum likhenge yahan par SP dash = SP square base ka square plus perpendicular humara PP dash hoga jo ki pichli triangle ke case mein bhi wahi tha.
[31:33]PP dash ka square upar power 1/2. Toh SP kitna hoga? SP kitna distance hai total yahan se leke yahan tak? Yahan se leke yahan tak A, yahan se leke yahan tak B. Toh SP kitna ho jaayega? A+B. Toh SP ki jagah humne (A+B) ka square put kar diya. PP dash humara X square hai, PP dash ka distance upar power 1/2. Ab hum SP dash ki value yahan put kar dete hain taaki humara CP dash nikal sake.
[31:51]Using four in third. Toh CP dash kitna aa jaayega? SP dash ki value put kar dijiye yeh wali jo humne fourth equation mein nikali hai, (A+B) ka whole square plus X square ki power 1/2 minus A yeh wala aise hi aayega. Yeh humari fifth equation aa gayi. Toh humne AP dash bhi find kar liya hai, CP dash bhi find kar liya hai. Ab humne dono ki values ko yahan put karna hai aur humein find karni hai maximum ki positions.
[32:13]Toh AP dash minus CP dash equal to 2n+1 lambda/2 jo hai. Humein AP dash and minus CP dash ki value put karni hai. Means AP dash, CP dash ki values second aur fourth equation se utha ke first mein daalni hai. Using fourth second and fourth in first. Toh AP dash ki value students yeh wali hai B square plus X square ki power 1/2. Woh aapne put kar deni hai. Minus CP dash ki value kaafi badi hai. Toh students, aapko yahan par brackets ka use karna padega curly brackets ka.
[32:41]Toh woh humare paas CP dash ki value yeh hai. Badi bracket start (A+B) ka whole square plus X square ki power 1/2 minus A, phir curly bracket close. Yeh poori ki poori jo hai woh CP dash ki value hai jo humne yahan put ki hai. Equal to 2n+1 lambda/2 jo hai poora last step tak aise hi chalta rahega. Humne sirf yeh wala part solve karna hai.
[33:10]Toh bade dhyan se solve karenge. Kaise karenge? Dekhiye, yahan se hum B square common lenge. Jab hum B square common lenge upar power 1/2 hai toh B square ki power 1/2 hai, usko agar hum bahar le toh yeh B ban ke bahar aayega. Two se two cancel ho jaayega. Toh B square ko jab hum power 1/2 se bahar lenge toh yeh B ban ke bahar aayega toh yahan bachega 1+X square aur B square yahan pe nahi tha toh uske upon mein aa jaayega.
[33:32]Minus yahan se hum A+B common lenge in dono terms se. Toh A+B ka whole square jab hum common lenge toh upar power 1/2 hai. Jaise wahan B square jab humne bahar liya tha toh B ban ke bahar aaya tha. Toh agar hum is bracket mein se (A+B) ki power 2 bahar lete hain toh upar 1/2 hai toh yeh (A+B) ban ke bahar aayega. Toh yahan par 1 reh jaayega plus X square divided by (A+B) ka whole square niche aa jaayega yahan par kyunki yahan par (A+B) ka square nahi tha woh upon mein aa jaayega yahan par.
[33:57]Aur yeh minus minus ho jaayega plus, ab curly bracket jo hain woh humne hata di hain. Equal to mein same chalega. Toh students, ab hum use karenge binomial theorem. Actually humare paas jo yeh distances hain X, B, A+B yeh bahut small distances hain. Yeh B bahut kam hai, A bahut kam hai, X bahut kam hai.
[34:14]Toh jab humare paas binomial expansion ke andar (1+X) ki power N ka jab hum binomial expansion karte hain toh humare paas woh yeh nikal ke aata hai, (1+NX). Toh agar N ki X ki value small hai, X ki value agar small hai toh humare paas toh hum isko hum (1+NX) likh sakte hain. Toh students, small powers small values of X ke liye, ya aap keh sakte hain small values of N ke liye kyunki yahan par N jo hai woh half hai.
[34:36](1+X) ki power N can be written as (1+NX). Isko bolte hain binomial expansion. Aage bahut saari terms hoti hain X square wali, X cube wali, woh hum neglect kar dete hain kyunki yahan par X jo hai woh bahut kam hai. Toh agar X bahut kam hai toh agar aap X square nikalenge toh aur bhi kam aayega. Kyunki aap X ko 0.1 maan lijiye, usko square kijiye toh 0.01, uska cube kijiye 0.001. Toh aise powers badhte rahenge toh uski values jo hain woh kam hoti rehti hain.
[35:03]Toh hum (1+X) ki power N ko hum aise likh sakte hain (1+NX). Toh humare paas 1+1+X humara yeh hai ki power N humari 1/2 hai. Toh usko likh sakte hain 1+NX. N kya hai humara 1/2 toh 1/2 yahan par aa jaayega. Yahan par humne 1/2 laga diya, 1/2 aur X square by B square humara aise hi rahega.
[35:24]Yeh wala formula humne use kar liya binomial theorem wala, jab X small ho toh hum yeh wala formula apply kar sakte hain. Minus (A+B) again yahan par binomial expansion lagayenge (1+X), X humara yeh poora term hai ki power N. Toh yeh hota hai (1+NX). N humara 1/2 aage aa gaya aur X humara yeh hai toh aise humne isko aise likh diya, plus A aur equal to mein same rahega. Ab hum brackets open karte hain.
[36:03]Jaise humne B aise hi likh diya. Yeh ek B se B cancel kar dete hain toh yahan bachega X square divided by 2B. Minus (A+B) ko jab 1 se karenge toh (A+B) aa jaayega. Minus iske saath jab hum 1 ko (A+B) se karenge toh yahan par (A+B) aa jaayega. X square divided by 2 (A+B) ka square aise hi rahega, plus A aur equal to mein same rahega. Ab hum cancel kar dete hain A se A kaat dete hain, B se B kaat dete hain. Toh students, yahan se agar hum X square by 2 common lein, X square by 2 toh humare paas bachega 1 upon B minus X square by 2 bahar chala gaya toh bachega -1 upon (A+B) equal to C.
[36:31]Toh iske aage ke steps jo hain aur kaise hum is expression se aage X ki value find karenge jo humein dega positions of maxima aur positions of maxima wale result ko use karke hum positions of minima bhi find karenge aur yeh sab dekhenge hum next section ke andar. Toh students, last section mein hum yahan tak nikal chuke the X square by 2 bracket start 1 by B minus 1 upon (A+B) equal to 2n+1 lambda/2.
[36:54]Ab hum yahan se X ki value nikalni hai taaki hum maxima ki positions nikal sakein. Toh uske liye humein is bracket ka LCM lena padega. LCM aa jaayega B (A+B), (A+B) yahan chala jaayega, B 1 se multiply ho jaayega. Toh aapke paas yeh aa jaayega aur B se B cancel ho jaayega toh aapke paas bachega X square A divided by 2B (A+B) equal to 2n+1 lambda/2.
[37:13]Aur yahan se cross multiply karke aap X square ki value nikal sakte hain. Cross multiply karke jab aap karenge toh 2 se 2 cancel ho jaayega toh B (A+B) upar chala jaayega aur yeh wala A jo hai woh niche aa jaayega yahan par A aayega aur yahan par lambda aayega. Toh yeh X square ki value aayi hai toh agar aap X ki value nikalna chahte hain toh aapko yahan par square root karna padega. Toh yeh maxima ki positions aa chuki hain.
[37:34]In-in positions ke upar maxima banega jahan par N ki values jo hain woh 0,1,2,3 and so on hain. Means jo maxima ki positions hain that is directly proportional to square root of odd natural number that is 2n+1. Toh yahan se aapko yahan tak karna hai maxima ki positions ke liye. Isi prakar se minima ki condition kya hoti hai diffraction ke andar ki path difference should be integral multiple of lambda.
[37:56]Aur path difference humne solve karke yeh find kiya tha. X square A divided by 2B (A+B). Agar yeh path difference ko agar hum N lambda ke barabar rakh dein jo ki destructive interference ki condition hoti hai. In-in positions par aapko minima dekhne ko milta hai. Agar path difference interfering beams ke beech mein integral multiple of lambda hai toh aapko minima dekhne ko milta hai. Toh path difference jo humne nikala tha solve karke wahi path difference aap yahan likhenge equal to N lambda likh denge.
[38:20]Toh cross multiply karenge toh aapko X square ki value mil jaayegi. X square ki value aa jaayegi cross multiply karke 2B (A+B) N lambda divided by N aur X ki value agar aap nikalna chahte hain toh yahan par square root aa jaayega aur yahan par aap minima likh denge kyunki yeh minima ki positions nikli hain.
[38:35]Toh minimum jo positions hongi that is directly proportional to square root of N jahan par N ki values jo hain woh 0,1,2,3 and so on ho sakti hain. Toh yahan tak karne ke baad, baki ka jo portion hai is topic ka aur jo intensity distribution diagram hoga is topic ka woh hum dekhenge agle section ke andar.
[38:52]Toh students, finally hum aa chuke hain apne last section of this topic pe jismein hum discuss karenge intensity distribution kaisa milne wala hai humein screen ke upar. Yeh hai humari screen thi. Is screen ko hum do parts mein divide kar sakte hain, pehla part P se upar wala jisko hum bolte hain illuminated region aur doosra part hai P se niche wala jisko hum bolte hain geometrical shadow region.
[39:11]Jo humara illuminated region hoga jo ki P se above hoga wahan par humein alternate dark and bright bands dekhne ko milenge. Woh bhi decreasing intensity ke aur unequal width ke dekhne ko milenge. Aur P se niche aapko thoda sa illumination milega aur jo illumination hoga woh ekdam bahut tezi se decrease hoga aur ultimately P se thoda sa niche ja ke aapko jo hai complete darkness dekhne ko milne wali hai. Toh aap graphically isko kaise represent karenge? Yeh humara P mid point of the screen jo humne yeh screen li hai usko P represent kar raha hai.
[39:41]Y axis pe humne intensity li hai aur X axis pe humne distance liya hai P se upar ka aur P se niche ka portion yahan par show kiya hai. P se upar wala portion jo hai use illuminated region naam de diya aur P se niche wala portion jo hai woh geometrical shadow region hoga. Toh jaisa ki aap dekh sakte hain ki illuminated region mein aapko alternate dark alternate bright, dark, bright, dark, bright jo bands dekhne ko mil rahe hain.
[40:04]Yeh jo peaks hain, yeh bright bands ko show kar rahi hain aur jo dips hain, yeh wala, yeh wala, yeh wala, yahan par aapko minima milega. Yahan par aapko maxima milega, first, yahan par second maxima, third maxima, fourth maxima and so on aur unki intensity jo hai woh lagataar jo hai woh kam hoti jaayegi. Aur P se jaise-jaise distance badhta jaayega waise-waise unki intensity jo hai woh kam hoti chali jaayegi.
[40:23]Aur P se is side means geometrical shadow region mein intensity jo hai woh ekdam gradually fall karegi aur rapidly fall karegi aur bilkul ek point par aakar P se thoda sa niche ja kar aapko jo hai complete darkness milegi aur intensity jo hai bilkul zero ho jaayegi. Toh students, aap yahan tak karke aayenge exam mein diffraction at a straight edge wale topic ke andar toh aapko 100% jo hain poore marks milenge. Toh students, I hope aapko aaj ka lecture samajh aaya hoga. Milenge next video mein till then thank you so much.
