Answer:
3.6 m
Explanation:
[tex]\lambda_R = 650 \ nm\\\\\lambda_R = 650*10^{-9} m\\\\L \ should \ be \ minimum \\\\i.e \ 0.25 \ mm\\\\= 0.25 *10^{-3} m[/tex]
[tex]\lambda_R = 700 \ nm\\\\\lambda_R = 700*10^{-9} m\\\\[/tex]
Also
[tex]\beta = 1 \ mm \ fringe \ width[/tex]
[tex]D_{min} = \frac{\beta d}{\lambda}\\\\D_{min} = \frac{10^{-3}*0.25*10^{-3}}{700*10^{-9}}\\\\D_{min} = 3.57 \\D_{min} = 3.6 m[/tex]
Therefore, the minimum distance L you can place a screen from the double slit that will give you an interference pattern on the screen that you can accurately measure using an ordinary 30 cm (12 in) ruler. = 3.6 m
