Diffraction When 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 Slit 1 . 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).

.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

A slit of width a is illuminate by white light ,For 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 ° ? C Incident wave （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

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

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,….

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 .

Diffraction from a Circular Aperture 1 . 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 .

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 ,

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 .

Diffraction by a double slit 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 .

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

. 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,

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 :

(4) An application of diffraction grating :Figure Shows a simple grating spectroscope in which a Grating to determine the wavelength

X-Ray Diffraction 1 .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

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