Question

quick refresher
this refresher from 35.2 youngs interference experiment will help you get familiar with prerequisite concepts
in a youngs double - slit experiment, the fringes are 1.0 mm apart. the screen is 1.8 m from the double slit that has a separation of 1.0 mm. what is the wavelength of the light?
○ 56 nm
○ 1800 nm
○ 180 nm
○ 5600 nm
○ 560 nm

Explanation:

Step1: Recall the formula for fringe separation in Young's double - slit experiment

The formula for the fringe separation (\(\Delta y\)) in Young's double - slit experiment is \(\Delta y=\frac{\lambda L}{d}\), where \(\lambda\) is the wavelength of the light, \(L\) is the distance between the double - slit and the screen, and \(d\) is the separation between the two slits. We need to solve for \(\lambda\), so we can re - arrange the formula to \(\lambda=\frac{\Delta y\times d}{L}\).

Step2: Convert the units to SI units

• The fringe separation \(\Delta y = 1.0\space mm=1.0\times10^{- 3}\space m\) (since \(1\space mm = 10^{-3}\space m\)).
• The distance between the double - slit and the screen \(L = 1.8\space m\).
• The separation between the two slits \(d = 1.0\space mm = 1.0\times10^{-3}\space m\).

Step3: Substitute the values into the formula

Substitute \(\Delta y = 1.0\times10^{-3}\space m\), \(d = 1.0\times10^{-3}\space m\) and \(L = 1.8\space m\) into the formula \(\lambda=\frac{\Delta y\times d}{L}\).

\(\lambda=\frac{(1.0\times10^{-3}\space m)\times(1.0\times10^{-3}\space m)}{1.8\space m}\)

First, calculate the numerator: \((1.0\times10^{-3}\space m)\times(1.0\times10^{-3}\space m)=1.0\times10^{-6}\space m^{2}\)

Then, divide by the denominator: \(\lambda=\frac{1.0\times10^{-6}\space m^{2}}{1.8\space m}\approx5.56\times10^{-7}\space m\)

Convert the wavelength to nanometers. Since \(1\space m = 10^{9}\space nm\), we have \(\lambda = 5.56\times10^{-7}\space m\times10^{9}\space nm/m=556\space nm\approx560\space nm\)

Answer:

560 nm (the option with "560 nm")