Question

Orange light of wavelength $6000\times {10}^{–10}\text{m}$ illuminates a single slit of width $0.6\times {10}^{–4}\text{m}$. The maximum possible number of diffraction minima produced on both sides of the central maximum is

Answer 1

For obtaining secondary minima at a point, path difference should be integral multiple of wavelength

$\therefore d\sin \theta =n\lambda$

$\therefore \sin \theta =\frac{n\lambda }{d}$

For $n$ to be maximum, $\sin \theta =1$

$\therefore n=\frac{d}{\lambda }=\frac{6\times {10}^{-5}}{6\times {10}^{-7}}=100$

Total number of minima on one side $=99$

$\therefore$ Total number of minima $=198$.