Question

A fringe width of $6mm$ was produced for two slits separated by $1mm$ apart. The screen is placed $10m$ away. The wavelength of light used is $\text{'}x\text{'}nm$. The value of $\text{'}x\text{'}$ to the nearest integer is ________.

Answer 1

Step 1. Given data:

Screen distance, $D=10m$

Fringe width,

$\begin{array}{rcl}\beta & =& 6mm\\ & =& 6\times {10}^{-3}m\end{array}$

Slit separation,

$\begin{array}{rcl}d& =& 1mm\\ & =& 1\times {10}^{-3}m\end{array}$

Wavelength, $x=?$

Step 2. Finding the wavelength of the light:

$\beta =\frac{xD}{d}$

Putting values in the above formula

$\Rightarrow$$\begin{array}{rcl}6\times {10}^{-3}& =& \frac{x\times 10}{1\times {10}^{-3}}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}x& =& 6\times {10}^{-7}m\end{array}$

$\begin{array}{rcl}\therefore x& =& 600nm\end{array}$

Thus, the wavelength of light used is, $\mathit{x}$ is $\mathbf{600}\mathbf{}\mathit{n}\mathit{m}$.

Hence, the value of $\mathit{x}\mathbf{=}\mathbf{}\mathbf{600}$.