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- 1. Vivian Tsang LO9 14153143
- 2. Light coming from two slits form a pattern on the screen shown. This pattern is made up of “fringes” that are light and dark The light fringes indicate areas of constructive interference The dark fringes indicate areas of destructive interference
- 3. At a center point, the distance two waves travel are equivalent and so constructive interference occurs at point P The two waves arrive in phase
- 4. The lower wave travels one wavelength farther than the upper wave. The waves arrive in phase at the point indicated This results in a bright fringe Constructive interference occurs d
- 5. Constructive interference Path difference must be zero or an integral multiple of the wavelength Variables θm= =the angle shown in the previous slide λ=wavelength d= distance between the two slits dsin(θm)=mλ m=….-3, -2, -1, 0, 1, 2, 3……etc m=order number of the fringes (ie: m= -/+2 is the second order bright fringe
- 6. Assuming paths of travelling light are parallel because the distance between the screens is large L>>d ymbright L
- 7. If we approximate tan(x)~sin(x) y~Lsin(θ) for small angles Therefore… sin(θ)=y/L Spacing between two bright fringes can be calculated by: ymbright=m λL/d As indicated on the previous slide: ymbright=the vertical distance between center of screen to the bright fringe L= the horizontal distance between the slits and the screen
- 8. The lower wave travels half a wavelength farther than the upper wave. The waves arrive out of phase at the point indicated This results in a dark fringe Destructive interference occurs
- 9. Therefore… sin(θ)=y/L Spacing between two dark fringes can be calculated by: ymdark=((m+0.5)λL)/d m=….-3, -2, -1, 0, 1, 2, 3……etc
- 10. If the distance between two sources are increased, the spacing between adjacent bright fringes will… Increase Decrease Stay the same
- 11. Decrease Reason: ymbright=m λL/d If d increases that will increase the denominator of the equation. This in turn will decrease the value of y Therefore, the spacing between adjacent bright fringes decreases
- 12. A light source consists of two types of gasses that emit light at 550nm and 400nm. The source is used in a double slit experiment. Note: assume that these sources do not interfere with each other What is the lowest order 550nm bright fringe that will fall on a 400nm dark fringe? What are their corresponding orders?
- 13. λ1=a= 550nm (bright) y1= (m1aL)/d) λ2=b=400nm (dark) y2= ((m2+0.5)bL)/d
- 14. y1=y2 (m1aL)/d) = ((m2+0.5)bL)/d (m1a) = ((m2+0.5)b) (m1550) = ((m2+0.5)400) 550m1=400m2+200 400(m2-m1)=150m1-200 Therefore, 150m1-200 must be a multiple of 400
- 15. Thus this will generate two answers: 400(m2-m1)=150m1-200 The lowest m1 where the RHS can be a multiple of 400 requires m1 to be =4 150(4)-200=400 Plugging m1=4 into the equation above gives us m2=5 Another interpretation leads us to conclude that m1 is =-4 150(-4)-200=-800 Plugging m1=-4 into the equation above gives us m2=6 Therefore: m1=4, m2=5 m1=-4, m2=6