Title 7. P2C7. ভৌত আলোকবিজ্ঞান (With Solve).pdf ✅

†f.Z Av‡jvKweÁvb  Engineering Practice Sheet ............................................................................................................ 3 12. `ywU mgeZ©b dvwj‡K mgvšÍiv‡j Ggbfv‡e ivLv n‡j †hb wØZxqwUi Av‡jvK Aÿ cÖ_gwUi Av‡jvK A‡ÿi mv‡_ 60 †Kv‡Y _v‡K| †Kv‡bv AmgewZ©Z Av‡jv‡K G m3⁄4vq GK cÖvšÍ w`‡q cvVv‡j Aci cÖv‡šÍi Av‡jvi ZxeaZv AmgewZ©Z Av‡jvi KZ ̧Y n‡e? [BUET 04-05] mgvavb: I2 = I0 2 cos2 60 = I0 2      1 2 2 = I0 8  I2 I0 = 1 8 A_©vr mgewZ©Z Av‡jvi ZxeaZv AmgewZ©Z Av‡jvi ZxeaZvi 1 8 ̧Y n‡e| (Ans.) 13. GKwU miæ †iLvwQ`a Øviv d«bndvi AceZ©b m„wói Rb ̈ †jÝ n‡Z 2 m `~‡i c`©v ivLv n‡jv| †iLv wQ‡`ai cÖ ̄’ 0.2 mm n‡j †`Lv hvq †h †K›`axq D3⁄4¡j we›`yi Dfq cv‡k¦© 5 mm `~i‡Z¡ Aeg we›`y MwVZ nq| AvcwZZ Av‡jvi •`N© ̈ wbY©q K‡iv| [BUET 03-04] mgvavb: 5 mm = 5  10–3 m 2 m c`©v wPo  tan = 5  10–3 2   = tan–1     5  10–3 2 = 0.143 a sin = n   = a sin n = 0.2  10–3 sin (0.143) 1 = 5  10–7 m (Ans.) 14. GKwU `yB w ̄øU cixÿvq cÖ_g me©wb‡¤œi †K.wYK Ae ̄’vb 0.20| w ̄øU `ywU ga ̈Kvi `~iZ¡ wbY©q K‡iv| e ̈eüZ Av‡jvi Zi1⁄2‣`N© ̈ = 5700 A  | [BUET 02-03] mgvavb: 1g me©wb¤œ ev Ae‡gi Rb ̈, a sin =     2n – 1 2   a sin (0.20) =     2  1 – 1 2  5700  10–10  a = 8.16  10–5 m (Ans.) weMZ mv‡j KUET-G Avmv cÖkœvejx 1. GKwU AceZ©b †MÖwUs Gi cÖwZ †mw›UwgUv‡i 6000 wU †iLv Av‡Q, hvnvi gva ̈‡g †mvwWqvg Av‡jvi wØZxq Pi‡gi eY©vjx cvIqv hvq| 2 wU †mvwWqvg Av‡jvi Zi1⁄2‣`N© ̈ 5890 A  Ges 5896 A  n‡j G‡`i g‡a ̈ †K.wYK `~iZ¡ KZ? [KUET 19-20] mgvavb: Pi‡gi Rb ̈, 1 N sin = n   = sin–1 (Nn)  2 – 1 = sin–1 (Nn2) – sin–1 (Nn1) = sin–1 (2  6000  5896  10–8 ) – sin–1 (2  6000  5890  10–8 ) = 0.058 (Ans.) 2. (a) Av‡jvK Kx? (b) †h wewKi‡Yi Zi1⁄2‣`N© ̈ 1.75  10–4 cm Zvi wd«‡Kv‡qwÝ ev ̄ú›`b msL ̈v wbY©q K‡iv| Av‡jvi MwZ = 3.0  1010 cms–1 | [KUET 03-04] mgvavb: (a) Av‡jv GK cÖKvi Zwor‡P.¤^Kxq Zi1⁄2| (Ans.) (b) c = f  f = c  = 3  108 1.75  10–4 = 1.71  1014 Hz (Ans.) 3. 5200 A  Zi1⁄2‣`‡N© ̈i meyR Av‡jv GKwU m~2 wPo n‡Z Bqs Gi wØ-wPo G AvcwZZ n‡”Q| 200 cm `~‡i c`©vi Dci 10wU cwÆi `~iZ¡ 4 cm| wP‡oi e ̈eavb wbY©q K‡iv| [KUET 03-04] mgvavb: x = nD a  a = nD x = 10  5200  10–10  200  10–2 4  10–2 = 2.6  10–4 m (Ans.) weMZ mv‡j RUET-G Avmv cÖkœvejx 1. 1.62 cÖwZmiv1⁄4 wewkó Kuv‡Pi †cø‡U Av‡jvK iwk¥ AvcwZZ nq| hw` cÖwZdwjZ Ges cÖwZmwiZ iwk¥ G‡K Ac‡ii mv‡_ j¤^fv‡e Ae ̄’vb K‡i Z‡e AvcZb †Kv‡Yi gvb wbY©q Ki| [RUET 19-20] mgvavb: r i  90 q i = P = mgeZ©b †KvY tanP =  P = tan–1 () = tan–1 (1.62) = 58.31 (Ans.)