Single Slit Diffraction

In single slit diffraction, NCERT gives the condition for the dark fringes as $a\sin \theta =m\lambda$, where $a$ is the slit width, $\lambda$ the wavelength, and $m$ the order of the minimum. For the first minimum $m=1$, and for small angles $\sin \theta \approx \theta$, so $\theta \approx \frac{\lambda }{a}$. Substituting $\lambda =500\times 10^{-9}$ m and $a=0.1\times 10^{-3}=10^{-4}$ m gives $\theta \approx \frac{500\times 10^{-9}}{10^{-4}}=\frac{5\times 10^{-7}}{10^{-4}}=5\times 10^{-3}$ radian, or 0.005 radian. The value 0.05 radian is wrong because it misplaces the slit-width exponent by one order. The value 0.0025 radian is wrong because it halves the correct result, perhaps confusing the central maximum half-width incorrectly. The value 0.5 radian is wrong because it is far too large for such a narrow-wavelength-to-width ratio. As stated in NCERT Class 12, Chapter 10, the central maximum broadens as the slit narrows, since $\theta$ scales as $\lambda /a$. A magnitude check confirms that a slit two hundred times wider than the wavelength gives a small angle of a few milliradians, consistent with 0.005 radian.