Question

A beam of light of wavelength 600 nm from a distinct source falls on a single slit $1.0mm$ wide and the resulting diffraction pattern is observed on a screen 2 m away. The distance between the first dark fringes on either side of the central bright fringe is?

Answer 1

Hint : In single slit diffraction, the dark fringe is located at an integral multiple of the wavelength. The difference between the locations of the dark fringes on either side gives us the distance between them.

Formula used: In this solution we will be using the following formula;
$y = \dfrac{{m\lambda D}}{d}$ where $y$ is the location of the dark fringes from the central maximum, $m$ is the mth place of the particular dark fringe on either side of the central maximum. (given as $m = .... - 3, - 2, - 1,0,1,2,3,....$ ), $\lambda$ is the wavelength of the light, $D$ is the distance of the screen from the slit, and $d$ is the width of the single slit.

Complete step by step answer
In single slit diffraction, a light wave is incident on a single slit of a particular width, a diffraction pattern, quite similar but different to double slit, is obtained on a screen at a distance away. In this diffraction pattern, the location of the dark fringes is given as
$y = \dfrac{{m\lambda D}}{d}$ where $m$ is the place (integer position) of the particular dark fringe, $\lambda$ is the wavelength of the light, $D$ is the distance of the screen from the slit, and $d$ is the width of the single slit. $m$ takes the values …-3, -2, -1, 0, 1, 2, 3… depending on which side of the central maximum it is.
Hence, for the first dark fringe on one side, we have
$y = \dfrac{{\left( 1 \right)\lambda D}}{d}$ , inserting all known values, we get
$y = \dfrac{{600 \times {{10}^{ - 9}}\left( 2 \right)}}{{0.001}} = 0.0012m$
Similarly, for the bright fringe on the other side, the distance is simply
$y = \dfrac{{\left( { - 1} \right)\lambda D}}{d}$ , which by insertion of known values will give
$y = \dfrac{{ - 600 \times {{10}^{ - 9}}\left( 2 \right)}}{{0.001}} = - 0.0012m$
The distance between them is hence,
$\Delta y = 0.0012 - \left( { - 0.0012} \right) = 0.0024m$
$\therefore \Delta y = 2.4mm$ .

Note
Alternatively, since the location of the bright fringes are equal from on both sides from the central maximum, then the distance between then is simply twice the distance of one from the central maximum as in:
$\Delta y = 2\dfrac{{\lambda D}}{d}$
Hence, by insertion of values again
$\Delta y = 2\dfrac{{600 \times {{10}^{ - 9}}\left( 2 \right)}}{{0.001}} = 0.0024m$