A) \[\frac{(3{{D}_{1}}+2{{D}_{2}})\lambda }{d}\]
B) \[\frac{(2{{D}_{1}}+3{{D}_{2}})\lambda }{d}\]
C) \[\frac{(3{{D}_{2}}-3{{D}_{1}})\lambda }{d}\]
D) \[\frac{(3{{D}_{1}}-2{{D}_{2}})\lambda }{2d}\]
Correct Answer: A
Solution :
[a] The fringe width P is given by, \[\beta =\frac{D\lambda }{d}\] where \[D={{D}_{1}}+({{D}_{1}}+{{D}_{2}})+({{D}_{1}}+{{D}_{2}})=3{{D}_{1}}+2{{D}_{2}}\] \[\therefore \,\,\,\beta =\frac{(3{{D}_{1}}+2{{D}_{2}})\lambda }{d}\]
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