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| Diffraction |
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| Diffraction in sound waves and radio waves are readings observed as they have a relatively longer wavelength compared to light waves. |
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| In the above figure l << a, when the light (plane) waves pass through the slit and the barrier forms a sharp shadow. As the slit 'a' is made smaller, light falls out into the formerly shadow region. This enrichment of light into the geometric shadow region is called diffraction. Note that 'a' is the transverse dimension of the slit and the effect of diffraction can be ignored if the ration of a/l is large. In such a case, light appears to be trained in straight line paths and is represented as rays. Thus, 'Ray', or 'geometrical optic' as it is called, is actually a limiting case of wave optics or physical optics. |
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| Consider a parallel beam of light from a lens falling on a slit AB. As diffraction occurs, the pattern is focused on the screen XY with the help of lens L2. |
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| Each point on the unblocked portion of plane wavefront AB sends out secondary wavelets in all directions. The waves from points equidistant from the center C lying on the upper and lower half reach the point O with path difference being zero and hence reinforce each other producing a maximum intensity at the point 0. |
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| Consider a point 'P' on the screen. The intensity at P will depend on the path difference between the secondary waves emitted from the corresponding point of the wavefront. Consider a set of waves making an angle q with CO. Draw AN perpendicular to BK. The path difference between the secondary waves reaching P from |
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| A and B = BN = AB Sin q = a sin q |
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| where AB = a width of the aperture. |
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| If BN = a (i.e., path difference) the whole
wave front can be considered to be divided into two halves, i.e., CA and CB. The path
difference between A and C will be l/2 and
so for every point in the upper half AC, there is a corresponding point in the lower half CB for
which the path difference between the waves is l/2.
So P is a point of destructive interference. |
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| qn is the angle of diffraction of nth dark fringe from O. |
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If  the slit can be considered to be divided
into three equal slits. The path difference between secondary waves from
corresponding points in the first two parts will be l/2.
This gives rise to destructive interference whereas the secondary waves from the third part remain unused and so produce the first secondary maximum. |
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| where n = 1, 2, 3, ... |
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| This represents the general expression to find the position of the nth secondary maxima. Given below is the variation of intensity vs. Path difference (a sinq). |
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| The point O corresponds to the position of central bright maximum for which path difference asinq = l, 2l, 3l ..... are secondary maxima. |
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| The secondary maxima have a path difference asin q = 3l/2, 5l/2, 7l/2 ..... |
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| Notice that the intensities of secondary maxima diminishes as path difference increases . This can be seen below. |
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| Width of central maximum is the distance between first and secondary minimum on either side of O. |
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| If 'f ' is focal length of lens L2 held close to the slit, then
f»D and for small values of q. |
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| Width of central maximum is |
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| The above picture shows the diffraction fringe formed by light passing through holes of various sizes. |
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| This shows the diffraction fringes obtained at the edges of razor blades. |
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| When the source is a white light, the diffraction pattern is colored but the central fringe is always white. |
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| Other bands are colored. As the bandwidth is a to wavelength, red band has a higher wavelength and is broader than the violet band, whose wavelength is small. |
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| Note: Diffraction occurs due to interference of two wave trains coming from two different points of the same wavefront and so a special case of interference occurs. Also the width and intensities of fringes in a diffraction is different and not equally bright. In interference, width and intensities of fringes are all equal. |
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