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

A, B, D

Explanation

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)`