john parkinson © 1 john parkinson © 2 the bending of waves around corners - past an obstacle or...

John ParkinsonJohn Parkinson©©

1


2

THE BENDING OF WAVES AROUND CORNERS THE BENDING OF WAVES AROUND CORNERS

- PAST AN OBSTACLE OR THROUGH A GAP- PAST AN OBSTACLE OR THROUGH A GAP

Ripple Tank Image

Barrier

Wave Height

- Intensity

Single Slit Diffraction


3

Original wavefront

Secondary sources

Subsequent position of wavefront

HUYGEN’s CONSTRUCTION FOR A PLANE WAVEFRONT

“Every point on a wavefront acts as a source of secondary waves which travel with the speed of the wave. At some subsequent time the envelope of the secondary waves represents the new position of the wavefront.”


4

WAVES WAVES

wide gap

narrow gap

The central maximum is twice the width of the other

maxima

The central maximum is lower [less energy passes

through], but wider


5

WAVES

For first minimum

sin dOr for small

angles in radians

d

d = width of the gap


6

http://webphysics.ph.msstate.edu/javamirror/ipmj/java/slitdiffr/index.html

At this web site you can change the width of the slit and the

wavelength to see how theses factors affect the diffraction

pattern




7

The double slit pattern is superimposed on the much broader single slit diffraction pattern.

The bright central maximum is crossed by the double slit interference pattern, but the intensity still falls to zero where minima are predicted from single slit diffraction. The brightness of each bright fringe due to the double slit pattern will be “modulated” by the intensity envelope of the single slit

pattern.

Diffraction by a Double Slit

The double slit fringes are still in

the same placeSingle slit pattern

Double slit pattern


8

n=0

DIFFRACTION GRATINGEach slit effectively acts as a point source, emitting secondary wavelets, which add according to the principle of superposition

n=2n=1

n=1 corresponds to a path difference of one wavelength

n=2 corresponds to a path difference of two wavelengths

n=3 corresponds to a path difference of three wavelengths


9

Grating

Monochromatic light C

For light diffracted from adjacent slits to add constructively, the path difference = AC must be a whole number of wavelengths.

AC = AB sin and AB is the grating element = d

Hence d sin n d = grating element

metreperlinesofnumberd

1

A

B


10

DIFFRACTION GRATINGS WITH WHITE LIGHTDIFFRACTION GRATINGS WITH WHITE LIGHT

PRODUCE SPECTRA

400nm 500nm 600nm 700nm

UV IR


11

DIFFRACTION GRATING WITH WHITE LIGHTDIFFRACTION GRATING WITH WHITE LIGHT

nmnm lightvioletlightred 400,700

Hence in any order red light will be more diffracted than blue.

White Central maximum, n = 0

First Order maximum, n = 1

First Order maximum, n = 1

Second Order maximum, n = 2

Second Order maximum, n = 2

Several spectra will be seen, the number depending upon the value of d

A spectrum will result

nd sin

Grating

screen


12

n=0n=2 n=1n=3

grating

Note that higher orders, as with 2 and 3 here, can

overlap

Be aware that in the spectrum produced by a prism, it is the blue light which is most deviated


13

QUESTION 1

Given a grating with 400 lines/mm, how many orders of the entire visible spectrum (400 – 700 nm) can be produced?

Finding the spacing d of the “slits” (lines).

d = 1/400 = 2.5 x 10-3 mm = 2.5 x 10-6 m

d sin = n

sin = (n )/d = a maximum of 1 at 900

Why do we use 700 nm?

Hence there are 7 orders in all (white central order + 3 on each side)

6.310700

1105.2sin9

6

d

n


14

Question 2: Visible light includes wavelengths from approximately 400 nm (blue) to 700 nm (red). Find the angular width of the second order spectrum produced by a grating ruled with 400 lines/mm.

As before d = 2.5 x 10-6 m

For red light in the second order

For blue light in the second order

34.1 - 18.7 = 15.40

0

6

9

11

27.18

105.2

104002

Sin

d

nSin

blue

0

6

9

11

21.34

105.2

107002

Sin

d

nSin

red

john parkinson © 1 john parkinson © 2 the bending of waves around corners - past an obstacle or...

Documents