question_answer

The maximum intensity in Young?s double-slit experiment is Io. Distance between the slits is \[d\,\,=\,\,5\,\lambda \], where \[\lambda \] is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance\[D\,\,=\,\,10d\]?

A) \[\frac{{{I}_{0}}}{2}\]                                                        

B) \[\frac{3}{4}{{I}_{0}}\]

C) \[{{I}_{0}}\]                            

D) \[\frac{{{I}_{0}}}{4}\]

Correct Answer: A

Solution :

\[\,\Delta x\,\,at\,\,P=\frac{dx}{D}\,=\,\,\frac{{{d}^{2}}}{2D}\,\,=\,\,\frac{{{(5\lambda )}^{2}}}{2\times 10\times d}\] \[\,\Delta x\,\,=\,\,\frac{{{(5\lambda )}^{2}}}{2\times 10\times 5\lambda }\,\,=\,\,\frac{\pi }{4}\] \[\,\Delta \phi \,\,=\,\frac{2\pi }{\lambda }\,\times \Delta x=\frac{\pi }{2}\] \[{{I}_{0}}\,=\,4I\,\,\Rightarrow \,\,I=\frac{{{I}_{0}}}{4}\] \[{{I}_{net}}\,\,=\,I+I+2\,\,\sqrt{I}\,\,\sqrt{I}\,\cos \,\,\frac{\pi }{2}\,\,=\,\,2I\,=\,\,\frac{{{I}_{0}}}{2}\]