Answer on Question #5692, Physics, Optics Stte slient fetures of single slit Frunhofer diffrction pttern. The slit is verticl nd illuminted by point source. Also, obtin n expression for intensity distribution nd plot it. Answer Frunhofer diffrction dels with the limiting cses where the light pproching the diffrcting object is prllel nd monochromtic, nd where the imge plne is t distnce lrge compred to the size of the diffrcting object. Fig.1 Let source of monochromtic light be incident on slit of finite width, s shown in Fig. 1. We will put point source of light in the focus of converging lens. Then fter refrction in the lens prllel bems will be obtined (see Fig.2). Fig.2 In diffrction of Frunhofer type, ll rys pssing through the slit re pproximtely prllel. In ddition, ech portion of the slit will ct s source of light wves
ccording to Huygens s principle. For simplicity we divide the slit into two hlves. At the first minimum, ech ry from the upper hlf will be exctly 18 o out of phse with corresponding ry form the lower hlf. For exmple, suppose there re 1 point sources, with the first 5 in the lower hlf, nd 51 to 1 in the upper hlf. Source 1 nd source 51 re seprted by distnce /2 nd re out of phse with pth difference /2. Similr observtion pplies to source 2 nd source 52, s well s ny pir tht re distnce /2 prt. Thus, the condition for the first minimum is / 2sin / 2. The rgument cn be generlized to show tht destructive interference will occur when where m. sin m (1) Fig.3 We cn use phsors to determine the light intensity distribution for single-slit diffrction pttern. Imgine slit divided into lrge number of smll zones, ech of width y s shown t right. Ech zone cts s source of coherent rdition, nd ech contributes n incrementl electric field of mgnitude E t some point on the screen. We obtin the totl electric field mgnitude E t point on the screen by summing the contributions from ll the zones. The light intensity t this point is proportionl to the squre of the mgnitude of the electric field. The incrementl electric field mgnitudes between djcent zones re out of phse with one nother by n mount, where the phse difference is relted to the pth difference ysin between djcent zones by n expression given by n rgument similr to tht leding to wht we did with interference (Eq.(2)). 2 ysin (2)
Fig.4 To find the mgnitude of the totl electric field on the screen t ny ngle, we sum the incrementl mgnitudes E due to ech zone. For smll vlues of, we cn ssume tht ll the E vlues re the sme. It is convenient to use phsor digrms for vrious ngles, s shown t right. When, ll phsors re ligned s in Fig.4, becuse ll the wves from the vrious zones re in phse. In this cse, the totl electric field t the center of the screen is E N E, where N is the number of zones. The resultnt mgnitude E t some smll ngle q is shown in Fig. 4, b, where ech phsor differs in phse from n djcent one by n mount. In this cse, E is the vector sum of the incrementl mgnitudes nd hence is given by the length of the chord. Therefore, E E. The totl phse difference between wves from the top nd bottom portions of the slit is 2 2 N Ny sin sin (3) where N y is the width of the slit. As increses, the chin of phsors eventully forms the closed pth shown in Fig.4,c. At this point, the vector sum is zero, nd so E, corresponding to the first minimum on the screen. Noting tht N 2 in this sitution, we see from the eqution derived bove tht 2 2 sindrk sindrk Tht is, the first minimum in the diffrction pttern occurs where sin / drk ; At lrger vlues of q, the spirl chin of phsors tightens. For exmple, Fig. 4,d represents the sitution corresponding to the second mximum, which occurs when 36 18 54. The second minimum (two complete circles, not shown) corresponds to 72, which stisfies the condition 2. sin drk
Fig.5 We cn obtin the totl electric-field mgnitude E nd light intensity I t ny point on the screen by considering the limiting cse in which y becomes infinitesiml ( dy ) nd N pproches infinity. In this limit, the phsor chins shown previously become the curve shown t right. The rc length of the curve is E becuse it is the sum of the mgnitudes of the phsors (which is the totl electric field mgnitude t the center of the screen). From Fig.5, we see tht t some ngle, the resultnt electric field mgnitude E on the screen is equl to the chord length. From the tringle contining the ngle /2, we see tht sin 2 E /2 (4) where is the rdius of curvture. But the rc length E is equl to the product, where is mesured in rdins. Combine this informtion with the previous expression to write n expression for E s function of E nd E sin / 2 2 2 / 2 E 2sin 2 sin E (5) Fig.6 Becuse the resultnt light intensity I (see Fig.6) t point on the screen is proportionl to the squre of the mgnitude E, we find tht 2 2 sin / 2 sin sin / I Imx Imx / 2 sin / (6)