Question

In a YDSE experiment with slit seperation of $2mm$ and slit to screen seperation of $1m$, a beam of light consisting of two wavelengths $6500\stackrel{0}{A}$ and $5200\stackrel{0}{A}$ is used. Minimum distance (in mm) from central maximum, where bright fringes due to both wavelengths coincide is (Write up to two digits after the decimal point.)

Answer 1

$y=\frac{n\lambda D}{d}$
Let $n^{th}$ of 6500 $A^{\circ }$ & $m^{th}$ of 5200 $A^{\circ }$coincide.
$y_n=y_m\Rightarrow n\times 6500\frac{D}{d}=m\times 5200\frac{D}{d}$
$5n=4m$
So, $n=4$ and $m=5$
$y_n=\frac{4\times 6500\times 1\times {10}^{-10}}{2\times {10}^{-3}}=2\times 65\times {10}^{-5}m$
$=130\times {10}^{-2}mm=1.30mm$