Answer:
As shown in Fig. the bright fringes \[{{B}_{1}}\] and \[{{B}_{2}}\] on either side of O coincide with \[{{S}_{1}}\,and\,{{S}_{2}}\]respectively. Clearly, - \[\beta =\frac{d}{2}\] As \[\beta =\frac{D\lambda }{d}\] \[\therefore \]\[\frac{d}{2}=\frac{D\lambda }{d}\] or \[\lambda =\frac{{{d}^{2}}}{2D}\].
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