Question

(A) Young's double-slit experiment is performed with 560-nm light and a distance of 2.00 m between the slits and the screen. The tenth interference minimum is observed 7.30 mm from the central maximum. Determine the spacing of the slits (in mm).

(B) What If? What are the smallest and largest wavelengths of visible light that will also produce interference minima at this location? (Give your answers, in nm, to at least three significant figures. Assume the visible light spectrum ranges from 400 nm to 700 nm.)

smallest wavelength _______nm

largest wavelength________nm

Answer 1

• In young `s double slit experiment the position of the maxima and minima is given as x d / D =[ { (2m+1) / 2 } * wavelength ] ;; For mth dark  fringe    xd / D = m* wavelength   ;; For mth bright fringe solution -​​​​​​​   ​​​​​​​ given x = position of the fringe = 7.30 mm   d = spacing of the two slit   = unknown = ?       D = Distance between slit and screen = 2 m Wavelength = 560 nm   using the formula for m = 10 and condition of dark fringe we get -       ( 7.30 * 10-3 * d) / 2 = [{(2*10 + 1) /2 } * 560 * 10 -9 ] d = 1.611 mm

1. now for minima at the same position i.e. at x = 7.30 mm and d = 1.611mm , D = 2 m using formula written above for dark fringes we have Wave length =( 7.30 * 10 -3 ​​​​​​​ * 1.611 * 10 -3) / (2m+1) wavelength =( 11856.96 nm ) / (2m+1)

• Now for m= 0 ;; wavelength = 11856.96 nm ; not in visible range
• for m = 1 ;;;; wavelength = 11856.96 / 3 = 3952.32  nm which is not in the visible range
• and simillarly for values of m = 2,3,4,5,6, and 7 ;; wave length is not in visible range
• and for m= 8 ;; wavelength = 11856.96 / 17 = 697.4682 nm ;; it will be the largest wavelength   for m= 14 ;; wavelength = 11856.96 / 29 = 408.860 nm ;;; it is the smallest wavelength
• thus answer for part ( a) = 1.611nm
• and answer for part (b ) = smallest wavelength - 408.860 nm = largest wavelength = 697.468 nm .