Question

A narrow slit S transmitting light of wavelength λis placed a distance d above a large plane mirror, as shown in the figure (17-E1). The light coming directly from the slit and that coming after the reflection interfere at a screen ∑ 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?
Figure

Answer 1

Intensity above Mirror is ZERO

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.  

b) Here, 2d = equivalent slit separation D = Distance between slit and screen.  

• We know that for a bright fringe, $\Delta x$ = $\frac {y \times 2d}{D}= n \lambda$

But as there is a phase reversal of $\frac {\lambda}{2}$

$\frac {y \times 2d}{D} + \frac {\lambda}{2} = n \lambda$

Hence,  

$\frac {y \times 2d}{D} = n \lambda - \frac {\lambda}{2}$

or $y = \frac {\lambda D}{4d}$