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Obtain the formulas angular width and linear width of central maximum. |
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Answer» SOLUTION :Width of central maximum is the distance between the first minimum on either side of the central maximum. Suppose, width of slit a is very SMALL (in the order of mm) and the distance between slit and screen is D (in the order of meter). This is shown in figure.  For a minimum of one side of the central maximum on the screen `theta=(lamda)/(a)` where `theta` is half the width of angular width of central maximum. `:.` Angular width of central maximum `2theta=(2lamda)/(a)` `2theta=(arc(beta_(0)))/("RADIUS"(D))` `:.beta_(0)=Dxx2theta` `=(2Dlamda)/(a)""....(1)""[:.theta=(lamda)/(a)]` Angle of `(n+1)^(TH)` order minimum `beta_(0)` is width of linear central maximum. It is minimum distance between the central maximum distance. Linear width of secondary maximum : The angular width in .n. means the least angular difference in n and (n + 1). `:.` The angle of least angular in .n., `theta_(n)=(nlamda)/(a)` Angle of `(n+1)^(th)` order minimum `theta_(n+1)=((n+1)lamda)/(Q)` `:.` Angular width of `n^(th)` secondary maximum, `theta_(n+1)-theta_(n)=(n+1)(lamda)/(a)-(nlamda)/(a)` `=(lamda)/(a)` `:.` The linear widht of `n^(th)` secondary maximum =angular width `xx` D `beta=(lamda)/(a)xxD""......(2)` `:.(beta_(0))/(beta)=2""` [ `:.` Ratio of equation (1) and (2)] `:.beta_(0)=2beta` |
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