The half angular width of the central bright maximum in Frauhoffer diffraction pattern, for a slit of width 1.2 × 10-3 mm illuminated by monochromatic light of wavelength 600 nm is :
Answer & explanation
Correct answer: option 2
As we know that in franhaufer diffraction
$sin\theta = \frac{\lambda}{a}$
where θ is the half angular width of the central maximum.
Given that $\lambda = 6\times 10^{-7}m , a = 1.2\times 10^{-6}m$
$\Rightarrow sin\theta = \frac{ 6\times 10^{-7}}{1.2 \times 10^{-6}} = 0.5 = 30^o$
$\theta = \frac{\pi}{6}$