# What is the width of the central bright fringe?

A light beam whose wavelength $\lambda$ is 550nm is passing through a single slit having a width of the slits equals to 0.4mm and hits a screen that is placed 2m away from the slit.

This question aims to find the width of the central bright fringe of the light that is passing through a slit and incident on a screen.

Single Slit Diffraction is the pattern that is developed when monochromatic light  with a constant wavelength $\lambda$ passes through a small opening of the size $a$ resultantly developing a Constructive and Destructive Interference which results in a bright fringe and a dark spot (minimum), respectively, which is represented by the following equation:

$a\ \frac{y_1}{D}=m\ \lambda$

Where:

$y_1=$ Distance between Central Fringe Center and dark spot

$D=$ Distance between Slit and Screen

$m=$ Order Destructive Interference

Central Bright Fringe is defined as the fringe that is brightest and largest and followed by smaller and lighter fringes on both sides. Its width is calculated by putting $m=1$ in the above equation:

$a\ \frac{y_1}{D}=(1)\ \lambda$

$y_1=\frac{\lambda D}{a}$

Since $y_1$ is the distance between the center of the Central fringe to the dark spot on one side, so the total width of the Central Bright Fringe is calculated by multiplying it by $2$ for both sides:

$y=2\frac{\lambda D}{a}$

Given that:

Wavelength of the light beam $\lambda=550nm=550\times{10}^{-9}m$

Size of the slit $a=0.4mm=0.4\times{10}^{-3}m$

Distance between Slit and Screen $D=2m$

We know that the Distance between Central Fringe Center and the dark spot is calculated as per the following formula:

$y_1=\frac{\lambda D}{a}$

Substituting the given values in above equation, we get:

$y_1=\frac{(550\times{10}^{-9}m)\times(2m)}{(0.4\times{10}^{-3}m)}$

$y_1=0.00275m$

$y_1=2.75\times{10}^{-3}m$

Since $y_1$ is the distance between the center of the Central fringe to the dark spot on one side, so the total width of the Central Bright Fringe is calculated by multiplying it by $2$ for both sides:

$y\ =\ 2\frac{\lambda D}{a}$

$y\ =\ 2(2.75\times{10}^{-3}m)$

$y\ =\ 5.5\times{10}^{-3}m$

## Numerical Result

The width of the central bright fringe after passing through a slit and incident on a screen is:

$y=\ \ 5.5\times{10}^{-3}m$

## Example

Light passes through a slit and incident on a screen having a central bright fringe pattern similar to that of electrons or red light (wavelength in vacuum $=661nm$). Calculate the velocity of the electrons if the distance between slit and screen remains the same and its magnitude is large in comparison to the slit size.

Solution

Wavelength of Electrons $\lambda=661\ nm=\ 661\times{10}^{-9}m$

We know that as per the relation for de Broglie wavelength of the electron, the wavelength of electrons depends on the momentum $p$ they carry as per the following:

$p={m}_e\times v$

So the wavelength of electrons is expresses as:

$\lambda=\frac{h}{p}$

$\lambda=\frac{h}{m_e\times v}$

By rearranging the equation:

$v=\frac{h}{m_e\times\lambda}$

Where:

$h=$ Plank’s Constant $=\ 6.63\times{10}^{-34}\ \frac{kgm^2}{s}$

$m_e=$ Mass of Electron $=\ 9.11\times{10}^{-31}kg$

$v=$ Velocity of Electron

$v=\frac{\left(6.63\times{10}^{-34}\ \dfrac{kgm^2}{s}\right)}{(9.11\times{10}^{-31}\ kg)\times(661\times{10}^{-9\ }m)}$

$v\ =\ 1.1\times{10}^3\ \frac{m}{s}$

Hence, the velocity of electron $v\ =\ 1.1\times{10}^3\dfrac{m}{s}$.