Question

Diffraction pattern from a single slit of width $0.25\, mm$ is observed with light of wavelength $5890\,\mathring{A}$. Angular separation between first order minimum and third order maximum, falling on the same side, is

• A. $5.89 × 10^{-3}$ rad
• B. $5.89 × 10^{-7}$ rad
• C. $5.89 × 10^{-10}$ rad
• D. $5.89 × 10^{-4}$ rad

Answer 1

Answer: (A)

In single slit diffraction pattern, positions of secondary maxima and minima are given by,
$\theta_{max}=\pm\left(2n+1\right) \frac{\lambda}{2d}$ and $\theta_{min}=\pm n \frac{\lambda }{d}$, respectively.
$\therefore \theta_{1\,min}=\frac{\lambda }{d}$ and $\theta_{3\,max}=\frac{7\lambda}{2d}$
Angular separation $= \theta_{3\,max}- \theta_{1\,min}=\frac{7\lambda }{2d}-\frac{\lambda}{d}$
$=\frac{5\lambda }{2d}=\frac{5}{2}\times\frac{5890\times10^{-10}}{0.25\times10^{-3}}=5.89\times10^{-3}$ rad.