Question

Question 3: Calibrating the Double Slit You are attempting to perform a double slit experiment (see Figure 2): the distance between slits is 0.1 mm, the distance from the barrier to the screen is L = 2 m, and the distance from the coherent) laser light source to the barrier is t = 0.4 m. You are using red light, with a wavelength = 660 nm. You are worried that you have not perfectly centred the laser halfway between the slits at y=0); rather, it may be slightly displaced in the direction indicated by the arrows. 0 i IW 1 Figure 2: Double slit experiment setup (not to scale). a) Indeed, when you measure the interference pattern at the screen, you find that the central maximum of constructive interference has been shifted down from the exact centre to y = -0.4 mm. By how much was the laser source displaced from the centre, and in what direction? b) In the situation from part (a), what is the phase difference between light at the two slits?

Answer 1

Solution:

displaced position TS - S displaced Source: W. original LOR position (Central maximum predicted i Y = -0.4 mm Coriginal Central maxima) 2 - ادا Now let the source has been shifted am. I am l=0.4m wao. 1x 103m so before attacking on the shit there is a path difference of sight of two slits so the path difference between the light of & stits. S, and se is 4, - OS, -os, By Gemetry 05 - Sa+w] + 6² m os : (a. wat After entering the right whent it reaches toward the point of Central maxima , at y = -0.4mm 4 - Szk' - S, k' Se k = 104x63 w jr - s, k' = [(4 x 163 ) ) 4 22

total path difference 4 = 4 + 4 Phase difference due to this path difference so 2 1 - 2T (4,+A,) Central maxima condition Cosso-) 1. Note Central maxima condition Intersity maxima 2 7 4 = 174 I = I + I + 2) I, I, loss I = Imax 47 when [cass=o [A + = 1 / 2 Is Imax - Int Iz Cos =0 central maxima py scaunel - Sa-] * [][6400 m n o - Jest wy**(] - * So here -3 w = 0.1mm - 0.1 x 10 m L=2m. 3 = 0.05 x 107 4- (0.45 x 6 3 ) ² +4 - ) (0.35x103) + 4 - ) 4.000000 203 - ) 4.000000 123 2.675 4108 m

4. + 2.675 x 108 = 1 4 + 2-675 x 108 - 600x109 1- 6oomm - 600 x 10 A- A. + 2.675 * 108. 150 x169 4, +2:675 x 16-8. 15x168 where 4. Je+ (at weg? - Slaug you can easily see that there a, w in mm and e is mm and l»(a + ) or l »[a-u) 4,- 1(21+4 (4 ) - 1 (1+1 (4.7") 1=1 (na lata) - 1 (1+4 (1991 4, = e. 4.2. - So, we have 4 + 2.675 x 108 = 15 x 158 4, - (15- 2-675) x168 4 = 12.325 x 10 m

4 is the path the light of difference between slit s, and se SO 4 = 12.325 x 10 m. wa - 12.325 x 108m w-o. 1x 103 m. e=aum af 4.93 x 104 m a = 0.493 x 103 m 0.493 mm a 0.5 mm 6. path light at the difference between slits (s, and se) A = 12-325 x 158 m. so, phase difference - 21.4. - 2x TT X 12-325 x 168 600 + 109 di = 1.2906