Question

In a Lloyd's mirror experiment, a light wave emitted directly by the sources $S$ interferes with the reflected light from the mirror. The screen is $1m$ away from the source $S$. The size of the fringe width is $0.25mm$. The source is moved $0.6mm$ above the initial position, the fringe width decreases by $1.5$ times. If the light wave used is in the wavelength $2.4\times {10}^{-x}$, find the value of $x$.

Answer 1

$\beta =\frac{\lambda D}{d}=0.25\times {10}^{-3}$ or $\lambda D=\frac{d}{4}\times {10}^{-3}$
Case $\left(ii\right)$ $\beta =\frac{\lambda D}{d+1.2\times {10}^{-3}}$ $=\frac{0.25\times {10}^{-3}d}{1.5}=\frac{{10}^{-3}}{6}$
or $\lambda D$ $=\frac{{10}^{-3}d}{6}+\frac{1.2\times {10}^{-3}\times {10}^{-3}}{6}$
$\lambda =\frac{d\times {10}^{-3}}{4D}=\frac{0.6\times {10}^{-3}\times {10}^{-3}}{1.0}=0.6\mu m$
$\frac{d}{4}=\frac{d}{6}+\frac{1.2\times {10}^{-3}}{6}$ or $d=2.4mm$.