![]() This defines Rayleigh’s resolution criterion. They would appear as one when the two central maxima overlap. ![]() When two objects placed at a distance from each other are separated by an angular separation θ, the diffraction patterns of the two objects will overlap each other. Such an image viewed from an optical device is calculated using Rayleigh’s criterion. The minimum distances between images must be such that the central maximum of the first image lies on the first minimum of the second and vice versa. The resolving power of a lens used in a telescope can be determined by its ability to differentiate two lines or points in an object. When a telescope is used to view two stars located close to each other the telescope’s resolving power will depend on its capacity to resolve the images of the two stars. The images of two close-lying objects appear distinct and separate when viewed from the device. In other words, resolving power changes in inverse proportion to the distance between the two objects to be resolved when viewed from an optical instrument. The resolving power of an optical instrument is defined as the capacity of the instrument to distinguish between two objects that are close together and produce distinct images of the two objects. The centre maximum’s width is just twice this amount.Īngular width of central maximum = 2θ = 2λ a Resolving Power The position of the minima determined by y (as measured from the screen’s centre) is, The maxima are located between the minima, and the width of the central maximum is equal to the distance between the 1 st order minima on both sides of the screen. Similarly, we may divide the slit into 2n parts for the nth fringe and utilise the following condition: We may divide the slit into four equal portions of a ⁄ 4 and use the same rationale for the next fringe. As each ray originating from a point has a counterpart that produces destructive interference, there is destructive interference at θ = sin −1(λ ⁄ a). There is another beam at a distance of a ⁄ 2 that can create destructive interference for a ray coming from any point in the slit. The path difference must create destructive interference for a dark fringe the path difference must be out of phase by λ ⁄ 2. In a minute, we’ll discover how important this method is. Any arbitrary pair of rays separated by a ⁄ 2 can be taken into account. ![]() Role of Mahatma Gandhi in Freedom StruggleĪny number of ray pairs that start at a distance of a ⁄ 2 from one another, such as the bottom two rays in the diagram, can be considered. ![]()
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