Diffraction Applications

Physics 202Professor Lee

CarknerLecture 26

PAL #25

The first side pattern is between the m=1 and m=2 diffraction minima:

a sin = and a sin = 2

sin = /a and sin = 2 /a

sin 1 = 650 X 10-9 / 0.08 X 10-3 = 8.125 X 10-3

sin 2 = (2)(650 X 10-9)/0.08 X 10-3 =1.625 X 10-2

PAL #25 What interference maxima are between

the two angles?

m1 = d sin 1 / and m2 = d sin 2 /m1 = (0.25 X 10-3)(8.125 X 10-3)/650 X 10-9=

m2 = (0.25 X 10-3)(1.625 X 10-2)/650 X 10-9= We should see 3 bright fringes (m = 4,5,6)

in the first side diffraction envelope

PAL #25

Middle interference fringe is m = 5 sin = (5)()/(d) =

= (a/ sin = [()(0.08 X 10-3 ) /(650 X 10-

9 )] (0.013) = 5.026 rad = (d/) sin =

I = Imcos2 (sin /)2 = Im (1)(0.0358) = 0.036 Im

PAL #25 Screen is 2 meters away, what is at point 4.3

cm from the center?

Diffraction pattern y/D = m/a, m = ya/D = (4.3X10-2)(0.08X10-3)/(2)

(650X10-9) = 2.65

Diffraction Gratings

If light of 2 different wavelengths passes through, each will produce a maxima, but they will tend to blur together

This makes lines from different wavelengths easier to distinguish

A system with large N is called a diffraction grating and is useful for spectroscopy

Maxima From Grating

Location of Lines

d sin = m where d is the distance between any two slits (or

rulings) on the grating

For polychromatic light, each maxima is composed of many narrow lines (one for each wavelength the incident light is composed of)

Elemental Lines

When electrons move between these energy levels, they can produce light at a specific wavelength

The pattern of spectral lines can identify the

element

Spectroscope

This will produce a series of orders, each order containing lines (maxima) over a range of wavelengths

The wavelength of a line corresponds to its position angle

We measure with a optical scope mounted on a vernier position scale Can also take an image of the pattern

Using Spectroscopy

We want to be able to resolve lines that are close together

How can we achieve this?

Line Width The narrower the lines, the easier to

resolve lines that are closely spaced in wavelength

hw = /(Nd cos ) where N is the number of slits and d is

the distance between 2 slits

Dispersion

D =

D = m / d cos For larger m and smaller d the

resulting spectra takes up more space

Resolving Power The most important property

of a grating is the resolving power, a measure of how well closely separated lines (in ) can be distinguished

R = av/

For example, a grating with R = 10000 could resolve 2 blue lines (= 450 nm) that were separated by 0.045 nm

Resolving Power of a Grating

R = Nm

Looking at higher orders helps to resolve lines

Spectral Type

The types of elements present in a star and the transitions they make depends on the temperature

Examples: Very cool stars (T~3000 K) can be identified by

the presence of titanium oxide which cannot exist at high temperatures

Next Time

Final exam Monday, 9-11 am, SC304 Bring pencil and calculator 4 equation sheets provided Covers 2/3 optics, 1/3 rest of

course

In a double slit diffraction pattern, what happens to the number of interference maxima in the first side pattern if you double the width of each slit?

a) Increasesb) Decreasesc) Stays the same

In a double slit diffraction pattern, what happens to the number of interference maxima in the first side pattern if you double the distance between the slits?

a) Increasesb) Decreasesc) Stays the same

In a double slit diffraction pattern, what happens to the number of interference maxima in the first side pattern if you double the wavelength?

a) Increasesb) Decreasesc) Stays the same

In a double slit diffraction pattern, what could to do to maximize the number of fringes in the central pattern?

a) Increase a, increase db) Decrease a, decrease dc) Increase a, decrease dd) Decrease a, increase de) You can’t change the number of

fringes in the central pattern