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Interference effects for continuous sources: i)Light bends around corners. ii)“Shadows” fill in iii)“Parallel” beams always spread iv)Limits of resolution.

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Presentation on theme: "Interference effects for continuous sources: i)Light bends around corners. ii)“Shadows” fill in iii)“Parallel” beams always spread iv)Limits of resolution."— Presentation transcript:

1 Interference effects for continuous sources: i)Light bends around corners. ii)“Shadows” fill in iii)“Parallel” beams always spread iv)Limits of resolution of microscopes and telescopes Diffraction Ch 38

2 Fraunhofer Diffraction: (easy math) Source, screen “ at infinity”, ~parallel rays eg.Laser & narrow slit. Fresnel Diffraction: (complicated math) Source distance, object size and screen distance all comparable. eg. SourceSteel Ball Screen Shadow Poisson Spot

3 Single Slit, width = a a θ Use Huygens’s idea: Treat the slit as a large number of point sources.  x = a sin θ is path difference between waves at two edges (Fraunhoffer diffraction)

4 Divide slit in half and add up rays in pairs: Ray (1) travels farther than ray (3): when it is ½ cycle behind -> Cancel Ray (2) travels farther than ray (4): when it is ½ cycle behind -> Cancel Ray (3) travels farther than ray (5): when it is ½ cycle behind -> Cancel 1) ANY two waves that originate at points a/2 away are out of phase by 180 o for some angle θ, and cancel. 2) waves from upper half interfere destructively with waves from the upper half a θ

5 By dividing the slit into four parts rather than two, we can similarly obtain: By dividing the slit into six parts rather than two, we can similarly obtain: Therefore, the general condition for destructive interference is: where m is the order number, for dark fringes

6 m = 3 m = 2 m = 1 m = -1 m = -2 m = -3 m=1.5 m=-1.5 θ The equation: Gives us the positions (θ) of the MINIMA !!! The maxima are approx half way between the minima.

7 Quiz: What would the pattern look like if the width of the slit was increased? A)The minima and maxima would be spaced very far apart B)No change, the pattern would look the same C)The minima and maxima would be spaced very closely together

8 Example 1: Light of λ=580nm is incident on a slit of width 0.30 mm. The observing screen is placed 2.0 m from the slit. a)Find the positions of the first dark fringes b)Find the width of the central bright fringe

9 Example 2: Light of λ=580nm is incident on a slit of width 0.30 mm. The observing screen is placed 2.0 m from the slit. a) Determine the width of the first order bright fringe. b) How does it compare to the width of the central max?

10 = 600 nm; central peak is 6cm wide on a screen 3m away. How wide is the slit? slit Example 3 L=3.0 m θ1θ1 θ1θ1 3 cm =y 1 3 cm

11 Intensity distribution for a single slit of width a Intensity What are the values of m and  ?

12 a sin θ = m λ  =m2  m = ±1, ±2, ±3, … ( but not m = 0) θ ∆y The intensity is a resultant of all the incremental electric fields from each part of the slit. These are out of phase by ∆β: (Since asinθ=mλ)

13 Example 4: Show that: a)the Minima happen when : a sinθ = m λ, m =±1, ±2… b)to what angles of β/2 does this correspond?

14 Intensity β/2 I1I1 I2I2 I2I2 I1I1 π -π-π -2π 2π2π -3π 3π3π

15 Example 5: Find the intensities of the first two secondary maxima for the single slit Fraunhofer diffraction pattern: (Hint: assume that this occurs midway between the minima.)

16 Notes: 1)Central maximum is twice as wide, and much brighter (~ 75% of light) 2)Secondary maxima get fainter as we move to higher orders m 3)Minima are at 4)Secondary maxima are approximately halfway between the minima. That is:


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