3.5.1 Single slit diffraction

Transcription

1 3.5.1 Single slit diffrction Wves pssing through single slit will lso diffrct nd produce n interference pttern. The reson for this is to do with the finite width of the slit. We will consider this lter. Tke look t the following simultion: Chnge the wvelength of the incident light. 1)! Wht hppens to the spcing of the bright fringes on the screen when the wvelength is incresed? Now chnge the slit width. 2)! Wht hppens to the spcing of bright fringes on the screen when the slit width is incresed? It is found tht the width of the centrl bright fringe is given by the expression: W = 2 where =wvelength, =slit to screen distnce, =slit width If we look t the intensity pttern for the light on the screen we get something like the following: W We cn see tht the centrl bright fringe hs width W. Subsequent bright fringes hve hlf the width of the centrl fringe. Note, too tht the intensity flls rpidly from centrl fringe to subsequent fringes. 3)! Find the width of the centrl bright fringe, when =1x10-5 m, =3m, nd the =450nm.

2 Let us return to the simultion bove. You will see tht the insted of finding W, n ngle cn be found by moving the sliders until the white rrows point to the drk fringes either side of the centrl bright fringe. W/2 We cn see how the ngle is relted to W. Therefore: And finlly: W 2 = tn W = 2 tn = 2 tn = 4)! In the simultion, select =550nm, =1500nm. Move the ngle slider until the rrows point to the first drk fringe. Record the ngle. Now use the formul bove to clculte the ngle. Show you working. You should get the sme nswer!

3 Huygens theory of wve propgtion To explin how wve fronts move forwrd, the utch scientist Christin Huygens suggested tht you should consider ech point on the wve front s source of secondry wvelets tht spred out in ll directions. It is the combintion of these smll wvelets tht produce n dvncing wve front. wve front time 4 time 3 time 2 time 1 Hve look t the following simultion: In this simultion you will see wve front being refrcted nd reflected t boundry between two medi. If you click you will see the Huygens explntion for these processes. Now consider wves pssing through gp. Wves diffrct round the edges. Stright hed ll the wves from ll points will rrive t distnt screen roughly in phse. However t n ngle, wves will rrive t the screen with phse difference becuse the pth length will be different for wves originting from different points. distnt screen

4 Consider the digrm bove. At certin ngle wves from the fr left of the slit will hve to trvel n extr wvelength more thn wves from the fr right of the slit. This mens they would be in phse. However, wves from the middle will hve to trvel hlf wvelength further nd so will be in ntiphse. If we dd up ll the wves from points ll the wy long the gp, we will dd wves for one whole cycle, i.e. with phse differences When we do this we end up with them ll cncelling ech other out destructive interference. At this point we would find the first drk fringe. W/2 For this sitution bove we know we should get destructive interference the first drk fringe) t this prticulr ngle, becuse wves originting from the lower prt of the slit hve to trvel n extr wvelength. The ngles re the sme. We cn write: nd: = sin W 2 = tn The smll ngle pproximtion sin = tn) cn be used here becuse is very lrge compred to. Therefore: W = 2

5 Hence: s shown t the top. W = 2 5)! If the 1 st drk fringe is found t n ngle, t wht ngle would you find the next drk fringe?