Answer:
Fringe width in Young's double slit experiment \[\beta =\frac{D\lambda }{d}\] In first case, \[{{\lambda }_{1}}=400\,nm=400\times {{10}^{-9}}m\] \[{{d}_{1}}=d\] [say] \[{{\beta }_{1}}=\frac{{{D}_{1}}{{\lambda }_{1}}}{{{d}_{1}}}\] In second case, \[{{\lambda }_{2}}=600\,nm=600\times {{10}^{-9}}m\] \[{{d}_{2}}=\frac{d}{2},\] \[{{\beta }_{2}}=\frac{{{D}_{2}}{{\lambda }_{2}}}{{{d}_{2}}}\] According to the question, \[{{\beta }_{1}}={{\beta }_{2}}\Rightarrow \frac{{{D}_{1}}{{\lambda }_{1}}}{{{d}_{1}}}=\frac{{{D}_{2}}{{\lambda }_{2}}}{{{d}_{2}}}\] or \[\frac{{{D}_{1}}}{{{D}_{2}}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}\times \frac{{{d}_{1}}}{{{d}_{2}}}=\left( \frac{600\times {{10}^{-9}}}{400\times {{10}^{-9}}} \right)\times \left( \frac{d}{\frac{d}{2}} \right)\] \[=2\times \frac{3}{2}=\frac{3}{1}\Rightarrow {{D}_{1}}:{{D}_{2}}=3:1\]
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