**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**.

The main concept behind this article is the **Single Slit Diffraction** **Patters**, **Destructive Interference**, and **Central Bright Fringe**.

**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:

\[y=2\frac{\lambda D}{a}\]

## Expert Answer

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** $v\ =\ 1.1\times{10}^3\dfrac{m}{s}$.