Question

A low-pressure sodium vapor lamp whose wavelength is 5.89 x 10−7 m passes through double-slits that are 6.7 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?

2.7 meters
3.0 meters
3.6 meters
4.2 meters

Answer 1

Answer:

3.6 meters

Explanation:

The formula that gives the position of the maxima on the distant screen for double-slit diffraction is

$y=\frac{m\lambda D}{d}$

where

m is the order of the maximum

$\lambda$ is the wavelength of the source

D is the distance of the screen

d is the distance between the slits

The distance between two consecutive fringes is therefore given by

$\Delta y = \frac{(m+1)\lambda D}{d}-\frac{m\lambda D}{d}=\frac{\lambda D}{d}$

where in this problem, we have:

$\lambda=5.89\cdot 10^{-7} m$ is the wavelength of the source

$d=6.7\cdot 10^{-4} m$ is the distance between the slits

$\Delta y=3.2\cdot 10^{-3} m$ is the distance between the fringes

Therefore, solving the equation for D, we find the distance to the screen:

$D=\frac{d\Delta y}{\lambda}=\frac{(6.7\cdot 10^{-4})(3.2\cdot 10^{-3})}{5.89\cdot 10^{-7}}=3.6 m$