Question

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 1

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

$\beta = \frac{\lambda L}{d}$

here we know that

$\beta = 3.2 \times 10^{-3}$

$\lambda = 5.89 \times 10^{-7} m$

$d = 2.0 \times 10^{-4} m$

now we have

$3.2 \times 10^{-3} = \frac{(5.89 \times 10^{-7})L}{2.0 \times 10^{-4}}$

$L = 1.1 m$

Answer 2

Answer:

it is 1.1 meters

Explanation: