Question

Light with a wavelength of 576 3. ×10−9 m passes through two slits and forms an interference pattern on a screen 8.85 m away. if the linear distance on the screen from the central fringe to the first bright fringe above it is 5.36 cm, what is the separation of the slits?

Answer 1

The correct formula for the double-slit experiment is given by
$y= \frac{n \lambda D}{d}$
where
y is the distance of the n-th maximum from the central fringe on the screen
n is the order of the maximum
D is the distance between the slits and the screen
d is the distance between the two slits

In our problem, we are given:
$\lambda=576.3 \cdot 10^{-9} m$, the wavelength of the light
$D=8.85m$, the distance of the screen from the slits
$y_1 = 5.36 cm = 0.0536 m$ is the distance of the first maximum (n=1) from the central fringe

Therefore, we can re-arrange the equation to calculate the separation between the slits:
$d= \frac{n \lambda D}{y_n}= \frac{(1)(576.3 \cdot 10^{-9} m)(8.85 m)}{0.0536 m}=9.52 \cdot 10^{-5} m$