Angular width of a central max. is $30^o$ when the slit is illuminated by light of wavelength 6000 Å. Then width of the slit will be approx.
Answer & explanation
Correct answer: option 2
$\text{Angular width } 2\theta = 30^o$
$sin 30^o = \frac{\lambda}{a}$
$a = \frac{\lambda}{sin30^o} = 2\lambda$
$ a = 12000A^o = 12\times 10^{-7}m$