 
                 
                InterviewSolution
 Saved Bookmarks
    				| 1. | A narrow slit S transmitting light of wavelength `lamda` is placed a distance d above a large plane mirror as shown in figure. The light coming directly from the slit and that coming after the reflection interference at a screen `sum` placed at a distance LD from the slit. a. What will be the intensity at a point just above the mirror. i.e., just above O? b. At what distance from O does the first maximum occur? ? | 
| Answer» Correct Answer - A::B::D a. Since there is a phase difference of `pi` between direct light and reflecting light the intensity just above the mirror will be zero.ltbr. B. Here 2nd= equivalent slit seperationn D=distance between slit and screen. We know for bright fringe `/_x=(yxx2d)/D=nlamda` But as thereis a phase reversal of `lamda/2` `rarr (yxx2d)D=lamda/2=nlamda` `rarr (yxx2d)/D=n lamda=lamda/2((n-1)/2)=lamda/2` `rarr y=(lamdaD)/(4D)` | |