Question

A beam of light consisting of wavelength 6000$\mathring{A}$ and 4500$\mathring{A}$ is used in a YDSE with D = 1m and d = 1mm. Find the least distance from central maxima, where bright fringes due to the two wavelengths coincide:

Answer 1

Suppose they melt at mth fringe of 1st and nth fringe of 2nd
$\therefore m{\beta }_1=n{\beta }_2$
$\Rightarrow m\times \frac{{\lambda }_{1}D}{d}=n\times \frac{{\lambda }_{2}D}{d}$
$\Rightarrow m\times 6000=n\times 4500$
m = 3 and n = 4 $\Rightarrow$ for minimum distance
$\therefore$ Minimum distance from
Central fridge = $m{\beta }_1=\frac{3\times 6000\times {10}^{-10}}{1\times {10}^{-3}}=1.8mm$