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

it is 1.1 meters

Step-by-step explanation:

Answer 2

Distance of the screen is approx 1.1 m

Step-by-step explanation:

As we know that fringe width on Young's double slit experiment is given as

`\beta = (\lambda L)/(d)`

here we know that

`\beta = 3.2 * 10^(-3)`

`\lambda = 5.89 * 10^(-7) m`

`d = 2.0 * 10^(-4) m`

now we have

`3.2 * 10^(-3) = ((5.89 * 10^(-7))L)/(2.0 * 10^(-4))`

`L = 1.1 m`