Question

Light of wavelength $6328\stackrel{o}{A}$ is incident normally on a slit having a width of $0.2mm$. The angular width of the central maximum measured from minimum to minimum of diffraction pattern on a screen $9.0$ meters away will be about

• A. $0.36degree$
• B. $0.18degree$
• C. $0.72degree$
• D. $0.09degree$

Answer 1

The correct option is A $0.36degree$

The angular width of central maxi. is
2$\theta$ = 2 $\frac{\lambda }{a}$ = $\frac{2\times 6328\times {10}^{-10}}{2\times {10}^{-4}}$ radian.
$=6328\times$ 10${}^{-6}\times \frac{180}{\pi }$

$degree={0.36}^o$