A) 4:1
B) 1:1
C) 1:4
D) 2:1
Correct Answer: A
Solution :
Given, \[\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{1}{2},\]\[\frac{{{d}_{1}}}{{{d}_{2}}}=\frac{2}{1}\] Fringe width in interference is given by, \[\beta =\frac{D\lambda }{d}\] Given \[{{\beta }_{1}}={{\beta }_{2}}\] \[\frac{{{D}_{1}}{{\lambda }_{1}}}{{{d}_{1}}}=\frac{{{D}_{2}}{{\lambda }_{2}}}{{{d}_{2}}}\] \[\frac{{{D}_{1}}}{{{D}_{2}}}=\frac{{{d}_{1}}}{{{d}_{2}}}\times \frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{2}{1}\times \frac{2}{1}=\frac{4}{1}\]
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