Question

The second-order dark fringe in a single-slit diffraction pattern is 1.40 mm from the center of the central maximum. Assuming the screen is 89.0 cm from a slit of width 0.710 mm and assuming monochromatic incident light, calculate the wavelength of the incident light.

Answer 1

We know, for single slit :

`y =( n\lambda L)/(a)\\\\\lambda = (ya)/(nL)` ...1)

`y = 1.4\ mm = 1.4 * 10^(-3)\ m`

n = 2

L = 89 cm = 0.89 m

`a=7.1* 10^(-4)\ m`

Putting all these in equation 1), we get :

`\lambda = (ya)/(nL)\\\\\lambda = (1.4* 10^(-3)* 7.1* 10^(-4))/(2* 0.89 )\\\\\lambda = 5.584 * 10^(-7)\ m`

Therefore, wavelength of the incident light is
`5.584 * 10^(-7)\ m` or 558.4 nm.

Hence, this is the required solution.