Question

A fraunhofer diffraction pattern is produced from a light source of wavelength 580 nm. the light goes through a single slit and onto a screen 1.0 m away. the first dark fringe is 5.0 mm from the central bright fringe. what is the slit width? (1 nm = 10-9 m). answer: 0.12 mm

Answer 1

for the dark fringe in diffraction we can say

$asin\theta = N\lambda$

here given that

n = 1 for first dark fringe

a = width of the slit

$\lambda = 580 nm = 580*10^{-9} m$

now for finding the angle it is given

for small angles we can write

$tan\theta = sin\theta = \frac{y}{L}$

$sin\theta = \frac{0.005}{1}$

now by above equation we can find

$a*0.005 = 1*580*10^{-9}$

$a = 1.16 * 10^{-4} m$

$a = 0.116 mm = 0.12 mm$

so width of the slit is 0.12 mm

Answer 2

Given, first dark fringe is 5mm away from central bright fringe.

Therefore,

Width of central bright fringe would be = 5 + 5 mm ( fringes form both

below and above the

centrall axis)

= 10mm

=> 2λD/a = 10mm

=> (2 x 580 x 10^-9 x 1)/a= 10 x 10^-3

=> a = 0.116mm ~ 0.12 mm