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Obtain the equation for resolving power of optical instrument - Physics

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प्रश्न

Obtain the equation for resolving power of optical instrument

दीर्घउत्तर

उत्तर

  1. For a single rectangular slit, the half-angle θ subtended by the spread of central maximum is given by the relation,
    a sin θ = λ
  2. Similar to a rectangular slit, when a circular aperture or opening forms an image of a point object, the image formed will not be a point but a diffraction pattern of concentric circles that become fainter while moving away from the centre. These are known as Airy’s discs. The circle of central maximum has the half angular spread given by the equation,
    a sin θ = 1.22 λ
  3. The numerical value 1.22 comes for central maximum formed by circular apertures.

    Airy’s discs
    For small angles, sin θ ≈ θ
    a θ =1.22 λ
    Rewriting further,
    θ = 1.22 λ/a and
    `"r"_0/"f" = (1.22 lambda)/"a"`
    `"r"_0 = (1.22lambda"f")/"a"`
  4. This equation is called spacial resolution.
  5. When two-point sources close to each another form image on the screen, the diffraction pattern of one point source can overlap with another and produce a blurred image. To obtain a good image of the two sources, the two-point sources must be resolved i.e., the point sources must be imaged in such a way that their images are sufficiently far apart that their diffraction patterns do not overlap.
  6. According to Rayleigh’s criterion, for two-point objects to be just resolved, the minimum distance between their diffraction images must be in such a way that the central maximum of one coincides with the first minimum of the other and vice versa. Such an image is said to be just resolved image of the object. Rayleigh’s criterion is said to be the limit of resolution.

    Rayleigh’s criterion
  7. According to Rayleigh’s criterion the two-point sources are said to be just resolved when the distance between the two maxima is at least r0. The angular resolution has a unit in radian (rad) and it is given by the equation,
    `theta = (1.22 lambda)/"a"`
  8. The ability of an optical instrument to separate or distinguish small or closely adjacent objects through the image formation is said to be the resolving power of the instrument.
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पाठ 7: Wave Optics - Evaluation [पृष्ठ १०४]

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सामाचीर कलवी Physics - Volume 1 and 2 [English] Class 12 TN Board
पाठ 7 Wave Optics
Evaluation | Q 11. | पृष्ठ १०४
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