Question

problem 24.02
the third - order bright fringe of 590 nm light is observed at an angle of 27° when the light falls on two narrow slits.
part a
how far apart are the slits?
express your answer to two significant figures and include the appropriate units
(d=) (\text{value}) (\text{units})

Explanation:

Step1: Recall Double - Slit Formula

The formula for the position of bright fringes in a double - slit experiment is \(d\sin\theta = m\lambda\), where \(d\) is the slit separation, \(\theta\) is the angle of the fringe, \(m\) is the order of the fringe, and \(\lambda\) is the wavelength of the light. We need to solve for \(d\), so we can re - arrange the formula to \(d=\frac{m\lambda}{\sin\theta}\).

Step2: Identify the Given Values

We are given that \(m = 3\) (third - order bright fringe), \(\lambda=590\space nm=590\times10^{-9}\space m\), and \(\theta = 27^{\circ}\).

Step3: Substitute the Values into the Formula

First, calculate \(\sin(27^{\circ})\). Using a calculator, \(\sin(27^{\circ})\approx0.454\). Then substitute \(m = 3\), \(\lambda = 590\times10^{-9}\space m\), and \(\sin\theta=0.454\) into the formula for \(d\):

\(d=\frac{3\times590\times10^{-9}\space m}{0.454}\)

\(d=\frac{1770\times10^{-9}\space m}{0.454}\)

\(d\approx3.9\times10^{-6}\space m = 3.9\space\mu m\) (since \(1\space\mu m = 10^{-6}\space m\))

Answer:

\(d = 3.9\space\mu m\) (or \(3.9\times 10^{-6}\space m\))