Light of wavelength 6000 Å falls on a single slit of width 0.1 ~mm...

Here, wavelength $\lambda=6000 Å=6000 \times 10^{-10} \mathrm{~m}$ Slit width $a=0.1 \mathrm{~mm}=0.1 \times 10^{-3} \mathrm{~m}$ In case of diffraction at single slit, the position of minima is given by $a \sin \theta_n=n \lambda$, where $n=1,2,3, \ldots$. second minimum, $n=2 \quad \therefore \quad a \sin \theta_2=2 \lambda$ For small angle $\sin \theta_2 \approx \theta_2$. $\therefore \theta_2=\frac{2 \lambda}{a}$ Substituting the values, we get $\therefore \theta_2=\frac{2 \times 6000 \times 10^{-10}}{0.1 \times 10^{-3}}=0.012$ radian