Question

A single slit diffraction pattern shown above. The light used to make the pattern has a wavelength of 561 nm and the screen is 2.6 m from the slit. If the slit has a width 3 mm, what is the width w of the central maximum in cm? 4.862x10^-4 Submit Answer Incorrect. Tries 2/20 Previous Tries

Answer 1

The first dark fringe is given by:

$sin\theta= \frac{\lambda}{w}$

where ${\lambda}$ is the wavelength of the light and w is the distance between the two slits.

Width of the central maxima is:

$2y= 2Ltan\theta$

where L is the distance between the slits and the screen

Hence,

$\theta= sin^{-1}\frac{\lambda}{w}= sin^{-1}\bigg(\frac{561 \times 10^{-9}}{3\times 10^{-3}}\bigg)\\ \theta= sin^{-1} (1.87 \times 10^{-4})= 0.0107^{\circ}\\ tan\theta= 1.87 \times 10^{-4}\\ 2y= 2\times 2.6 \times 1.87 \times 10^{-4}m 2y= 9.72 \times 10^{-4} m$