A low-pressure sodium vapor lamp whose wavelength is 5.89 x 10^−7 m passes through double-slits that are 2.0 x 10^−4 m apart and produces an interference pattern whose fringes are 3.2 x 10^−3 m apart on the screen. What is the distance to the screen?
3 meters
2.7 meters
1.6 meters
1.1 meters
its literally 1.1 meters I got it right
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]
Answer:
it is 1.1 meters
Explanation: