Discussion Forum

A light source, which emits two wavelengths $\lambda_{1}$= 400nm and $\lambda_{2}$=600 nm  . is used in Young's double-slit experiment. If recorded fringe widths for  $\lambda_{1}$  and $\lambda_{2}$ are  $\beta_{1}$  and $\beta_{2}$ and the number of fringes for them within a distance y on one side of the central  maximum are   $m_{1}$ and $m_{2}$ respectively, then

Answer:

Explanation:

Fringe width $\beta= \frac{\lambda D}{d}$   or $   \beta\propto\lambda $

 $\therefore$     $ \lambda_{2}$   >  $ \lambda_{1}$ 

 So     $\beta_{2}$  >  $\beta_{1}$

Number of fringes ina given width

   $m=\frac{y}{\beta}$    or   $m\propto\frac{1}{\beta}$

  $\Rightarrow$     $  m_{2}$   <  $m_{1}$   as    $\beta_{2}$ >   $\beta_{1}$

Distance of 3 rd maximum of  $\lambda_{2}$  from central maximum

       $=\frac{3\lambda_{2}D}{d}=\frac{1800D}{d}$

 Distance of 5 th minimum of  $\lambda_{1}$ from central maximum

                     $=\frac{9\lambda_{1}D}{2d}=\frac{1800D}{d}$

So,  3rd maximum of  $\lambda_{2}$ will overlap with 5th minimum of $\lambda_{1}$

 Angular seperation (or angular fringe width)

   $=\frac{\lambda}{d}\propto \lambda$

    $\Rightarrow $   aangular separation for $\lambda_{1}$  will be lesser.