Question

The orange light of wavelength$6000\times {10}^{-10}\mathrm{m}$ illuminates a single slit of width$0·6\times {10}^{-4}\mathrm{m}$. The maximum possible number of diffraction minima produced on both sides of the central maximum is?

Answer 1

Step 1. Given data:

1. The wavelength of orange light $\mathrm{\lambda }=6000\times {10}^{-10}\mathrm{m}$
2. Width of the slit $\mathrm{d}=0·6\times {10}^{-4}\mathrm{m}$

Step 2: Formula used:

The Bragg's equation for the diffraction minima is,

$\mathrm{dsin\theta }=\mathrm{n\lambda }$

[where $\mathrm{n}=$the number of diffraction minima, $\mathrm{\theta }=$angle of diffraction]

Step 3: Calculating the maximum possible number of diffraction minima:

As we know,

$\begin{array}{rcl}\mathrm{sin\theta }& =& \frac{\mathrm{n\lambda }}{\mathrm{d}}\end{array}$

For the central maxima, $\begin{array}{rcl}\mathrm{sin\theta }& =& 1\left(\mathrm{\theta }={90}^{\circ }\right)\end{array}$

So,

$\begin{array}{rcl}& \Rightarrow & \frac{\mathrm{n\lambda }}{\mathrm{d}}<1\\ & \Rightarrow & \mathrm{n}<\frac{\mathrm{d}}{\mathrm{\lambda }}\\ & \Rightarrow & \mathrm{n}<\frac{0·6\times {10}^{-4}\mathrm{m}}{6000\times {10}^{-10}\mathrm{m}}\\ & \Rightarrow & \mathrm{n}<100\end{array}$

On one side the number of diffraction minima would be - $99$

For both sides,

$99+99=198$

Thus, the maximum possible number of diffraction minima produced on both sides of central maxima is$198$.