A) \[6\,\Omega \]
B) \[3\,\Omega \]
C) \[20\,\Omega \]
D) \[3.33\times {{10}^{8}}\,m/s\]
Correct Answer: A
Solution :
In Young?s double slit experiment fringe width \[\left( W \right)\] is given by \[W=\frac{D\lambda }{d}\] Where \[D\] is the distance between the screen and source, d is distance between coherent sources and \[\lambda \] is wavelength of light employed. Given, \[{{\lambda }_{1}}=5000\,\overset{\circ }{\mathop{\text{A}}}\,,\,{{W}_{1}}=0.5\,mm\,\,{{\lambda }_{2}}=6000\overset{\circ }{\mathop{\text{A}}}\,.\] \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{5000}{6000}\] \[\Rightarrow \] \[\frac{0.5}{{{W}_{2}}}=\frac{5}{6}\] \[\Rightarrow \] \[{{W}_{2}}=\frac{3}{5}=0.6\,\,mm\] Note: Fringe width increases as wavelength of light increases.
You need to login to perform this action.
You will be redirected in
3 sec
