Question

Manufacturers of wire (and other objects of small dimension) sometimes use a laser to continually monitor the thickness of the product. The wire intercepts the laser beam, producing a diffraction pattern like that of a single slit of the same width as the wire diameter (Fig.$\mathbf{36}\mathbf{-}\mathbf{37}$). Suppose a helium – neon laser, of wavelength $\mathbf{632}\mathbf{.}\mathbf{8}\mathbf{}\mathbf{nm}$, illuminates a wire, and the diffraction pattern appears on a screen at distance $\mathit{L}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.}\mathbf{60}\mathbf{}\mathbf{m}$. If the desired wire diameter is $\mathbf{1}\mathbf{.}\mathbf{37}\mathbf{}\mathbf{mm}$, what is the observed distance between the two tenth-order minima (one on each side of the central maximum)?

Short answer

The distance between two tenth order diffraction minima on both side is $24\text{mm}$.

Step-by-step solution

Identification of given data

The wavelength of the laser is $\lambda =632.8\text{nm}$

The distance of screen from laser is $L=2.60\text{m}$.

The diameter of wire is $d=1.37\text{mm}$.

The order of tenth minima is $m=10$.

Concept used

The distance between two successive diffraction minima is called the fringe width. It varies with the wavelength, slit width and distance of screen from slit.

Determination of distance between two observed tenth diffraction minima on both side

The distance between two tenth order diffraction minima on both side is given as:

$w=\frac{2m\lambda L}{d}$

Substitute all the values in equation.

$$\begin{gathered} w=\frac{2\left(10\right)\left(632.8\text{nm}\right)\left(\frac{{10}^{-9}\text{m}}{1\text{nm}}\right)\left(2.60\text{m}\right)}{\left(1.37\text{mm}\right)\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)} \\ =24\times {10}^{-3}\text{m} \\ =\left(24\times {10}^{-3}\text{m}\right)\left(\frac{1\text{mm}}{{10}^{-3}\text{m}}\right) \\ =24\text{mm} \end{gathered}$$

Therefore, the distance between two tenth order diffraction minima on both side is $24\text{mm}$.