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 :

$y =\dfrac{ n\lambda L}{a}\\\\\lambda = \dfrac{ya}{nL}$       ...1)

$y = 1.4\ mm = 1.4 \times 10^{-3}\ m$

n = 2

L = 89 cm = 0.89 m

$a=7.1\times 10^{-4}\ m$

Putting all these in equation 1), we get :

$\lambda = \dfrac{ya}{nL}\\\\\lambda = \dfrac{1.4\times 10^{-3}\times 7.1\times 10^{-4}}{2\times 0.89 }\\\\\lambda = 5.584 \times 10^{-7}\ m$

Therefore, wavelength of the incident light is $5.584 \times 10^{-7}\ m$ or 558.4 nm.

Hence, this is the required solution.