Question

A narrow slit S transmitting light of wavelength $\lambda$ is placed a distance d above a large plane mirror as shown in figure (17-E1). The light coming directly from the slit

and that coming after the reflection interfere at a screen $\sum$ placed at a distance D 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 1

(a) Since, there is a phase difference of ${}^{\prime }{\pi }^{\prime }$ between direct light and reflecting light, the intensity just above the mirror will be zero.
(b) Here, 2nd = equivalent slit separation
D = Distance between slit and screen.
We know for bright fringe,
$△x\frac{y\times 2d}{D}=n\lambda$
But as there is a phase reversal of $\frac{\lambda }{2}$
$\Rightarrow \frac{y\times 2d}{D}=\frac{\lambda }{2}=n\lambda \Rightarrow \frac{y\times 2d}{D}=n\lambda =\frac{\lambda }{2}\left(\frac{n-1}{2}\right)=\frac{\lambda }{2}\Rightarrow y=\frac{yD}{4D}$