DiffractionWhen monochromatic light from a distance

source (or a laser ) passes through a narrow slit

and then intercepted by a viewing screen ,the

light produces on the screen a diffraction pattern

like that in figure this pattern consists of abroad

and intense (very bright )central maximum and

A number of

Narrower and less

Intense maxima

(called secondary or side maxima) to both side

In between the maxima are minima

Diffraction by a Single Slit1 . Let us consider how

plane waves of light of

wavelength are diffracted

by a single long narrow

slit of width a in an

otherwise opaque

screen B, as shown in

cross section in figure (a).

2 .We can justify the central bright fringe seen in

figure by noting that the waves from all points in

the slit travel about the same distance to reach

the center of the pattern and thus are in phase

there. As for the other bright fringes, we can say

only that they are approximately halfway

between adjacent dark fringes.

3 , to locate the first dark fringe at point p1 , we

first mentally divide the slit into two zones of

equal widths a/2 . Then we extend to p1 a light

ray r1 ,from the top point of top zone and a

light ray r 1 from the top point of the bottom zone. A central axis is drawn from the center of the slit to screen C, and

P 1 is located at an angle θto that axis. The first dark fringe can be located at . It means

2sin

2

a

sina

SAMPLE PROBLEM 1

A slit of width a is illuminate by white light ,For

What value of a will the first minimum for red

Light of λ=650 nm be at θ=15 ° ?

0p

1p

2p

B

CIncident wave

4

a

4

a

4

a

4

a

（ a）

SOLUTION : at the first minimum ,m=1 in Eq.

solving for a ,

wefind

(answer)

For the incident light to flare out that much ( )

The slit has to be very fine indeed ,amounting to

About four times the wavelength .Note that a

fine Human hair may be about 100 µm in

diameter

,,3,2,1,,sin mforma

mnmnmm

a

5.2251115sin

)650)(1(

sin

15

4 .To find the second dark fringes above

and below the central axis, we divide the slit

into four zones of equal widths , a/4 as

shown in above figure (a). We then extend rays

r1 , r 2, r 3, and r 4 from the top points of the

zones to point P2, the location of the second

dark fringe above the Central axis .the second

dark fringe can then be located at 2

sin4

a

We then extend rays r 1 , r 2, r 3, and r 4

from the top points of the zones to point P2,

the location of the second dark fringe above the

Central axis .the second dark fringe can then be

located at or

5. The dark fringes can be located with the

following general equation :

For m=1,2,3,….

2sin

4

a 2sin a

ma sin

Above equation is derived

for the case of .D>>a

However, it also apply if

we place a converging

lens between the slit and

the viewing screen and

then move the screen in

so that it coincides with

the focal plane of the

lens. Intensity in single

-Slit diffraction : we can

Prove the expression for the intensity I of the

pattern as ,where

And Im , is the greatest value of the intensity in

the pattern ,and it occurs at the central maximum

Where (θ=0) Figure shows plot of the intensity of

A single-slit diffraction pattern for three different

Slit width .

2)sin

(

mII sina

Diffraction from a Circular Aperture1 . Here we consider diffraction by a circular

aperture, through which

light can pass. Figure

shows the image of a

distant point source of light

formed on photographic

film placed in the focal

plane of a converging lens

(1) The image is not

a point But a circular disk surrounded

by several progressively fainter secondary rings.

(2) The first minimum for the diffraction pattern

of a circular aperture of diameter d is given by .

d

22.1sin

2 .Resolvability: In figure (b) the angular

separation of the two point sources is such that

the central maximum of the diffraction patter

of one source is centered on the firstminimum

of the diffraction pattern of the other, a

condition called Rayleigh’s criterion for

resolvability. Thus two objects that are barely

resolvable by this criterion must have an angular

separation , dR

22.1sin 1

SAMPLE PROBLEM 2

A converging lens ,32 mm in diameter and with a

Length f , of 24 cm ,is used to form images of

Objects . Considering the diffraction by the

Lens ,what angular separation must two distant

Point objects have to satisfy Rayleigh’s criterion ?

Assume that the wavelength of the light from the

Distant object is λ=550 nm

.

SOLUTION :From Eq we have

Of small angular separation ,it is desirable .these

When we wish to use a lens to resolve objects

These can be done either by increasing the

lens Diameter or by using light of a shorter

Wavelength .

dR

22.1sin 1

radm

m

dR5

3

91 1010.2

1032

)10550)(22.1(22.1sin

Diffraction by a double slit 1. In double-slit experiment ,we implicitly

Assumed that the slits were narrow compared to

The wavelength of the light illuminating them ;that

Is ,a<<λ. For such narrow slits ,the central

Minimum of the diffraction pattern of either slits

Covers the entire viewing screen ,Moreover ,the

interference of light from the two slits produces

bright fringes that all have approximately the

same intensity.

2. In practice with visible light, however, the

condition a<<λ is often not met. For relatively

wide slits, the interference of light from two slits

produces bright fringes that do not have the same

intensity. In fact, their intensity is modified by

the

diffraction of the light through each slit. See

the figures.

3. With diffraction effects taken into account, the

intensity of a double-slit interference pattern is

given by , in which

and .

22 )sin

)((cosmII

sin)(d

sin)(a

Diffraction Grating 1 . One of the most useful tools in the study of

Light and of objects that emit and absorb light is

The diffraction grating .some what like the

double-slit arrangement ,these devise has a

much Greater number N of slits ,often called

rulings, perhaps as many as several thousand

Per millimeter

2 . With monochromatic light on a diffraction

Grating if we gradually increase the number of

Slits From two to a large number N, The maxima

Are now narrow

(and so called lines )

They are

Separated by relatively

wide dark regions ,as

Shown in above figure

Look at the figure

An idealized

Diffraction grating

Containing five

Slits ,The scale is

Distorted for

Clarity

3 . The separation between rulings is called the

grating spacing. The maxima-lines are located

at , in which the

integers are then called the order numbers, and

the lines are called the zeroth-order line (the

central line, with m=0), the first-order line, the

second-order line, and so on.

4 . Width of the lines: (1) we measure the half-

width of the central line as the angle ΔθKλ

from the center of the line at θ=0 outward to

Where the line effectively ends and darkness,

,3,2,1,0,sin mformd

(1) we measure the half-width of the central line

as the angle ΔθKλ from the center of the line at

θ=0 outward to Where the line effectively ends

and darkness, Effectively begins with the first

minimum .

(2) the Half-width of the central line is given as(3)

(3) The half-width any other line is given as :

Ndhw

cosNdhw

(4) An application of diffraction grating :Figure

Shows a simple grating spectroscope in which a

Grating to determine the wavelength

X-Ray Diffraction1 .X rays are electromagnetic radiation whose

wavelengths are of the order of 1Ǻ . Figure

shows that x rays are produced when electrons

escaping from a heated filament F are

accelerated

by a potential

difference V and strike

a metal target T.

2 . Bragg’s law: .

Figure shows how the inter

planer spacing d can

be related to the unit

cell dimension :a0

,3,2,1sin2 mformd

Questions

Home work

Exercise and problems :

3E 8P

16P 39P

43E 66E

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