question_answer

In Young's double slit experiment, using light of wavelength 400 nm, interference fringes of width X are obtained. The wavelength of light is increased to 600 nm and the separation between the slits is halved. If we want the observed fringe width on the screen to be the same in two cases, then find the ratio of the distance between the screen and the plane of the interfering sources in the two arrangements.

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         \[\frac{{{D}_{1}}}{{{D}_{2}}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}\times \frac{{{d}_{1}}}{{{d}_{2}}}=\left( \frac{600\times {{10}^{-9}}}{400\times {{10}^{-9}}} \right)\times \left( \frac{d}{\frac{d}{2}} \right)\] \[=2\times \frac{3}{2}=\frac{3}{1}\Rightarrow {{D}_{1}}:{{D}_{2}}=3:1\]