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?
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