Double Slit Diffraction Physics 202 Professor Lee Carkner Lecture 27

PAL #26 Diffraction  Single slit diffraction, how bright is spot 5 cm from center?  = 680 nm, a = 0.25 mm, D = 5.5 m  Convert y =5 cm to   tan  = y/D,  = arctan (y/D) = 0.52 deg  Need to find  to find I   = (  a/ )sin  = 10.5 rad  I = I m (sin  /  ) 2 = I m  Nearest minima  What is m for our  ?  a sin  = m  m = (a sin  / = 3.33  Between 3 and 4, closer to 3

Double Slit Diffraction  In double slit interference we assumed a vanishingly narrow slit and got a pattern of equal sized (and equally bright) maxima and minima   In single slit diffraction we produced a wide, bright central maximum and weaker side maxima   The interference maxima are modulated in intensity by a broad diffraction envelope

Diffraction and Interference

Double Slit Pattern  The outer diffraction envelope is defined by: a sin  =m   Between two minima, instead of a broad diffraction maxima will be a pattern of interference fringes  d sin  = m  a,d and are properties of the set-up,  indicates a position on the screen and there are two separate m’s (one for the diffraction and one for the interference)

Patterns   e.g. You would expect the m = 5 interference maxima would be bright, but if it happens to fall on the m = 3 diffraction minima it will be dark   What you see at a certain angle , depends on both of the m’s  We can use the location of two adjacent diffraction minima (sequential diffraction m’s) to define a region in which may be several interference maxima   i.e. first define the diffraction envelope, then find what interference orders are inside

Diffraction Envelope

Diffraction Dependencies   For large (d) the interference fringes are narrower and closer together  For longer wavelengths the peaks are further apart   For solving diffraction/interference problems:   Can find the interference maxima with d sin  =m   There are two different m’s

Intensity   = (  a/ ) sin    = (  d/ ) sin   The combined intensity is: I = I m (cos 2  ) [(sin  /  ] 2

Diffraction Gratings  What happens when white light passes through a double slit?   But each maxima is very broad and they overlap a lot  If we increase the number of slits (N) to very large numbers (1000’s) the individual maxima (called lines) become narrow   A system with large N is called a diffraction grating  Used for spectroscopy, the determination of a materials properties through analysis of the light it emits at different wavelengths

Maxima From Grating

Location of Lines  The angular position of each line is given by: d sin  = m   The m=0 maxima is in the center, and is flanked by a broad minima and then the m=1 maxima etc.   Called an order

Using Gratings   Rather than a continuous spectrum of all colors, the gas only produces light at certain wavelength called spectral lines   By passing the light through a grating we can see these spectral lines and identify the element  Each element has a unique pattern of lines

Emission Lines of Hydrogen

Resolving Power and Dispersion  What do we want from our grating?   Gratings with a high resolving power (R) produce narrow lines R = Nm  D = m / (d cos  ) 

Next Time  Final Exam, Monday 9-11am  Study, PAL, notes, old tests  Bring pencil and calculator