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Unformatted text preview: Answers Question 1 (i) Vector Diagram If the phase of the light from the first slit is zero, the phase from second slit is sin 2 d = Adding the two waves with phase difference where  = x ft 2 , ( 29 ( 29 { } + = + + + = + + cos cos 2 ) cos( ) cos( 2 / ) 2 / cos( 2 ) cos( ) cos( a a a a a a This is a wave of amplitude cos 2 a A = and phase . From vector diagram, in isosceles triangle OPQ, sin 2 1 d = = ) 2 ( = NB and . cos 2 a A = Thus the sum of the two waves can be obtained by the addition of two vectors of amplitude a and angular directions 0 and . (ii) Each slit in diffraction grating produces a wave of amplitude a with phase 2 relative to previous slit wave. The vector diagram consists of a 'regular' polygon with sides of constant length a and with constant angles between adjacent sides. Let O be the centre of circumscribing circle passing through the vertices of the polygon. Then radial lines such as OS have length R and bisect the internal angles of the polygon. Figure 1.2. 1 Figure 1.2 d = = = S O T OTS T S O ^ ^ ^ and ) 180 ( 2 1 In the triangle TOS , for example sin 2 ) 2 / sin( 2 R R a = = as ) 2 ( = sin 2 a R = (1) As the polygon has N faces then: N N Z O T N Z O T 2 ) ( ^ ^ = = = Therefore in isosceles triangle TOZ, the amplitude of the resultant wave, TZ, is given by N R sin 2 . Hence form (1) this amplitude is sin sin N a Resultant phase is ( 29 ( 29 ) 1 ( 1 2 1 180 2 1 2 90 ^ ^ ^ =  = = N N N Z T O S T O S T Z (iii) 2 Intensity 2 2 2 sin sin N a I = N a sin sin 1 37C (iv) For the principle maxima ........ 2 1 where = = p p ' and ' ' ' 2 2 2 max + = = = = p a N N a I (v) Adjacent max. estimate I 1 : N N p N 2 3 i.e 2 3 2 , 1 sin 2 = = = = N p 2 does not give a maximum as can be observed from the graph.as can be observed from the graph....
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 Spring '11
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