Question

A narrow slit $S$ transmitting light of wavelength $\lambda$ is placed a distance $d$ above a large plane mirror as shown in fig $29.16$ (a). The light coming directly from the slit and that after reflection interference coverage at $P$ on the screen placed at a distance $D$ from the slit. What will be the intensity at a point just above $O$? What will be $x$ for which first maxima occurs?

Answer 1

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 for bright fringe,

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

But as there is a phase reversal of $\lambda /2$

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

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

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