Answer:
Distance of the screen is approx 1.1 m
Explanation:
As we know that fringe width on Young's double slit experiment is given as
[tex]\beta = \frac{\lambda L}{d}[/tex]
here we know that
[tex]\beta = 3.2 \times 10^{-3}[/tex]
[tex]\lambda = 5.89 \times 10^{-7} m[/tex]
[tex]d = 2.0 \times 10^{-4} m[/tex]
now we have
[tex]3.2 \times 10^{-3} = \frac{(5.89 \times 10^{-7})L}{2.0 \times 10^{-4}}[/tex]
[tex]L = 1.1 m[/tex]
