Question

A beam of light consisting of two wavelengths, $6500\stackrel{\circ }{A}$ and $5200\stackrel{\circ }{A}$, is used to obtain interference fringes in a Young's double slit experiment:
What is the least distance from the central maximum where the bright fringes due to both wavelengths coincide? The distance between the slits is $2mm$ and the distance between the plane of slits and the screen is $120cm$.

Answer 1

$LCM$ of $65$ and $52=260$

i.e., $\lambda =26000\mathring{A}$

$\beta =\frac{\lambda D}{d}=\frac{26000}{{10}^{10}}\times \frac{120}{100}\times \frac{1000}{2}$

$=\frac{156}{{10}^5}$m

$=1.56$mm