Question

lighr of wavelength 440 nm passws throufh a double slit, yieldinfa diffraction pattern whose graph of intensity I versus angular position

1) Light of wavelength 440 nm passes through a double slit, yielding a diffraction pattern whose graph of intensity I versus angular position is shown in the Figure below. Intensity (mW/cm") 0 5 8 (degrees) Calculate: (a) the slit width and (b) the slit separation. (e) Verify the displayed intensities of the m= 1 and m= 2 interference fringes.

Answer 1

diffraction minima bsino hod b = -g 1x 440*10 Sin 5° → Slit width m. ba 5.048 x10-6 b > 4th is missing Interference Interference maxima . dsino = mi 오 By -6 4.* 5.048 xlo m ds 1 da 2.02x10-5 slit Seperation La fo Sin²ß co²r p² lib sino Y= 2 And sino B: d

& m=2 for mal co Y = - e d sino = sino = 1 di mal rx 5.048 x 10 rbd Lig 2 x 10 ß= do radian 0.785 0.8/ fo Rosa 1,- lo sino.785 0.745 ) 5.67 mW = 0.81 x7 - 0.81lo cm² Amal -6 For ma2 211 y So 048 x16 5 ß- 20b व 2:02 x 10 ß? 1.57 Jadian sin? 1.57 1. Im=2 (1.57)? 0.411., IM= 2 = 0.41*7 = 2.87 xosh (ma