Q.Monochromatic electromagnetic radiation from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width if the wavelength is

[!TLDR] Central maximum full width $=2\lambda D/a$; solving $a=2\lambda D/(\text{width})$ for each given wavelength gives 0.4167 mm (visible), 41.67 mm (infrared), $4.167\times10^{-4} …

The central bright maximum of a single-slit diffraction pattern spans between the first minima on either side, at $\pm\lambda D/a$, so its total (full) width is $2\lambda D/a$ (Section 7.9.3). Given this width as 6.00 mm $=6\times10^{-3}$ m and screen distance $D=2.50$ m, rearranging for the slit width: $a=\dfrac{2\lambda D}{\text{width}}$.

1. Visible light, $\lambda=500\text{ nm}=5\times10^{-7}$ m: $a=\dfrac{2(5\times10^{-7})(2.5)}{6\times10^{-3}}=\dfrac{2.5\times10^{-6}}{6\times10^{-3}}=4.167\times10^{-4}\text{ m}=0.4167$ mm.
2. Infrared, $\lambda=50\ \mu\text{m}=5\times10^{-5}$ m: $a=\dfrac{2(5\times10^{-5})(2.5)}{6\times10^{-3}}=\dfrac{2.5\times10^{-4}}{6\times10^{-3}}=4.167\times10^{-2}\text{ m}=41.67$ mm.
3. X-rays, $\lambda=0.500\text{ nm}=5\times10^{-10}$ m: $a=\dfrac{2(5\times10^{-10})(2.5)}{6\times10^{-3}}=\dfrac{2.5\times10^{-9}}{6\times10^{-3}}=4.167\times10^{-7}\text{ m}=4.167\times10^{-4}$ mm. …

Use the central-maximum full-width formula (width $=2\lambda D/a$), rearrange to $a=2\lambda D/\text{width}$, and substitute each given wavelength in …

Using the HALF-width ($\lambda D/a$, the single-sided distance to one minimum) instead of the FULL central-maximum width ($2\lambda D/a$) that the question actually sp …