question_answer

The maximum intensity in Young's double slit experiment is \[{{I}_{0}}.\] Distance between the slits is J= \[d=5\lambda ,\]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 \[D=10\,d\]?

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

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

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

D)        \[\frac{{{I}_{0}}}{4}\]

Correct Answer: A

Solution :

[a] Suppose P is a point in front of one slit at which intensity is to be calculated from figure. It is clear that \[x=\frac{d}{2}.\]. Path difference between the waves reaching at P is \[\Delta =\frac{xd}{D}=\frac{\left( \frac{d}{2} \right)d}{10d}=\frac{d}{20}=\frac{5\lambda }{20}=\frac{\lambda }{4}\] Hence corresponding phase difference          \[\phi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}\] Resultant intensity at P               \[I={{I}_{\max }}{{\cos }^{2}}\frac{\phi }{2}\]             \[={{I}_{0}}{{\cos }^{2}}\left( \frac{\pi }{4} \right)=\frac{{{I}_{0}}}{2}\]