Question

Coherent light with wavelength 594 nm passes through two very narrow slits, and the interference pattern is observed on a screen a distance of 3.00 m from the slits. The first-order bright fringe is a distance of 4.84 mm from the center of the central bright fringe. For what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?

Answer 1

The position of bright fringes Y_m on screen in double slit experiment is given by following expressión

$y_m={m\lambdaD}{d}$

We can solve for d, clearing the variables, so

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

So making the substitution for λ= 594nm, D=3m, y_1=4.84 and m=1, we have

$d=\frac{594*10^{-9}*3*1}{4.84*10^{-3}}\\d=3.7*10^{-4}m$

Through this value we can find the position of dark fringes y_m on screen. The following expressión can approach better,

$y_m=(m+\frac{1}{2})\frac{\lambda D}{d}$

To solve the wavelength m is equal to 0,

$y_0=\frac{\lambda D}{2d}$

clearing for \lambda

$\lambda = \frac{2y_0 d}{D}$

Making the substitution for $y_0=4.84mm, d=3.7*10^{-4}, D=3m$

$\lambda=\frac{2(4.84*10^{-3})(3.7*10^{-4})}{3}\\\lambda=1.19 \mu m$

So we have that the wavelength required is $1.19 \mu m$