double slit interference phys 101 lo9
TRANSCRIPT
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
Some Basics
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
Interference Patterns
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
Bright Fringes
d
Constructive interferencePath difference must be zero or an integral
multiple of the wavelengthVariables
θm= =the angle shown in the previous slideλ=wavelengthd= distance between the two slits
dsin(θm)=mλm=….-3, -2, -1, 0, 1, 2, 3……etcm=order number of the fringes(ie: m= -/+2 is the second order bright fringe
General Form
Assuming paths of travelling light are parallel because the distance between the screens is large L>>d
Rationale ymbright
L
If we approximate tan(x)~sin(x) y~Lsin(θ) for small angles
Therefore…sin(θ)=y/LSpacing 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 fringeL= the horizontal distance between the slits and the
screen
Equations
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
Dark Fringes
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
Equations
If the distance between two sources are increased, the spacing between adjacent bright fringes will…
IncreaseDecreaseStay the same
Question 1: The Warm Up
Decrease
Reason:ymbright=m λL/dIf 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
Answer
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 otherWhat is the lowest order 550nm bright fringe
that will fall on a 400nm dark fringe?
What are their corresponding orders?
Question 2:
λ1=a= 550nm (bright)y1= (m1aL)/d)
Let’s work on the bright fringe:
Let’s work on the dark fringe:λ2=b=400nm (dark)y2= ((m2+0.5)bL)/d
y1=y2
(m1aL)/d) = ((m2+0.5)bL)/d (m1a) = ((m2+0.5)b)(m1550) = ((m2+0.5)400)550m1=400m2+200400(m2-m1)=150m1-200Therefore, 150m1-200 must be a multiple of 400
Equate the distances
Thus this will generate two answers:400(m2-m1)=150m1-200The lowest m1 where the RHS can be a multiple of 400
requires m1 to be =4150(4)-200=400
Plugging m1=4 into the equation above gives us m2=5Another interpretation leads us to conclude that m1 is =-4
150(-4)-200=-800Plugging m1=-4 into the equation above gives us m2=6
Therefore:m1=4, m2=5m1=-4, m2=6