Double slit interference phys 101 lo9

Vivian Tsang LO9 14153143

Double Slit Interference

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

Thanks for Watching!