Question

A pair of narrow slits is illuminated with light of wavelength λ= 566.1 nm. The resulting interference maxima are found to be separated by 0.90 mm on a screen 1.18 m from the slits. What is the separation of the slits? (mm)

Answer 1

PAGE No. for an interference pattern, the formula for the position of the bright fringe (maxima) given by from is x d -mi o 72 x= mdo d Where D = bistame & creen d= separation of slite a = wavele elength Separation between two consecutie maxima is called fringe width. Therefore, if Xm mt be the distances of the and the "[mu] the bright funges (manı'na) respectively from the central fringe o and (m+1 DA and mth then = MDA d Amar d Imatt - Home 3 separation befw the foreiges 08 distance between the manima (MH) DA m DA d d 24 MTI Dd d T PAPER

PAGE No. 2.......... Given that the separation between tuo and D = 1.18m, maxima is 0.98 mm 2 = 566.1 nm .. Unt - 2M - DA d m 0.98mm 1:18 × 566 l nm d -9 0.98 nm= 1.18 m X 566 11 x 10 d d = 1.18 566. 1x io9 x 18 m xn 0.98 x 19 9 9 9 d = 1.18 x 56601 m 0.98 9 dz 681.630 m 681.630 X10 de 681630. X10 mm 1m=10 nm d = 6 nm