question_answer

The maximum intensity in Young's double slit experiment is I0. Distance between the slits is d = 5l, where l 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)            I0    

D)            \[\frac{{{I}_{0}}}{4}\]

Correct Answer: A

Solution :

           Suppose P is a point infront 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                    \[\Delta =\frac{xd}{D}=\frac{\left( \frac{d}{2} \right)d}{10d}\]  \[=\frac{d}{20}=\frac{5\lambda }{20}=\frac{\lambda }{4}\] Hence corresponding phase difference \[\varphi =\frac{2\pi }{\lambda }\times \frac{\lambda }{4}=\frac{\pi }{2}\] Resultant intensity at P \[I={{I}_{\text{max}}}{{\cos }^{2}}\frac{\varphi }{2}\]\[={{I}_{0}}{{\cos }^{2}}\left( \frac{\pi }{4} \right)=\frac{{{I}_{0}}}{2}\]