In Young's double slit experiment a parallel light beam containing wavelength ${\lambda }_1=4000\stackrel{\circ }{A} \text{and} {\lambda }_2=5600\stackrel{\circ }{A}$ is incident on a diaphragm having two narrow slits. Separation between the slits is $d=2mm$. If distance between diaphragm and screen is $D=40 cm$, calculate the distance of first black line line from central bright fringe 

When a monochromatic light of wavelength $\lambda$ is used to obtain interference pattern in Young's double slit experiment, fringe width is given by $\omega =\frac{\lambda D}{d}$ where $D$ is distance of screen from slits and $d$ is distance between the slits.

Hence, fringe width for light of wavelength ${\lambda }_1, {\omega }_1=\frac{{\lambda }_{1}D}{d}$

and fringe width for light of wavelength ${\lambda }_2, {\omega }_2=\frac{{\lambda }_{2}D}{d}=112\mu m$${\omega }_1=80\mu m$

Since, the incident light beam has both the wavelengths ${\lambda }_1 \text{and} {\lambda }_2$, therefore interference patterns are formed on the screen for both the wavelength. A black line is formed at the position where dark fringes are formed for both of the wavelengths.

Let first black line be formed at distance $y$ from central bright fringe. Let at this position there bemth dark fringe of wavelength ${\lambda }_1$ and $n^{th}$ dark fringe of wavelength ${\lambda }_2$.

Distance of first black line, from central bright line,

$y=\left(m-\frac{1}{2}\right){\omega }_1=\left(n-\frac{1}{2}\right){\omega }_2$  …..(1)

or  $\frac{2m-1}{2n-1}\frac{{\omega }_2}{{\omega }_1}$  …..(2)

For first black line, $y$ should be minimum possible which corresponds to least possible integer values of $m$ and $n$.

Hence,   $\frac{2m-1}{2n-1}=\frac{7}{5} \text{or} m=4, n=3$

Position of first black line $y=\left(m-\frac{1}{2}\right){\omega }_1=280\mu m$