Q.What must be the ratio of the slit width to the wavelength for a single slit to have the first diffraction minimum at 45.0°? [Ans (as printed in the book): 1.274]

[!TLDR] The first-minimum condition $a\sin\theta=\lambda$ gives $a/\lambda=1/\sin45.0°=\sqrt2\approx1.414$; the book's own printed answer of 1.274 does not match this straightforward substitution, and the …

The condition for the FIRST diffraction minimum of a single slit is $a\sin\theta=\lambda$ (Section 7.9.3, with $n=1$). Rearranging directly for the required ratio: $\dfrac{a}{\lambda}=\dfrac{1}{\sin\theta}$. Substituting $\theta=45.0°$: $\sin(45.0°)=\dfrac{1}{\sqrt2}\approx0.7071$, so $\dfrac{a}{\lambda}=\dfrac{1}{0.7071}=\sqrt2\approx1.414$.

This is a direct, unambiguous one-line substitution into the textbook's own stated first-minimum formula, and the result $\sqrt2\approx1.414$ can be checked independently: at $a/\lambda=1.414$, $a\sin(45°)=1.414\times0.7071=1.000=\lambda$ exactly, confirming the first-minimum condition is genuinely satisfied. The book's own printed answer for this problem, 1.274, does NOT satisfy this check (it would correspond to a first-minimum angle of about 51.7°, not the 45.0° actually stated in the question) -- this looks like an error or OCR/typesetting corruption in the source answer key rather than a different intended method, sin …

Rearrange the single-slit first-minimum condition $a\sin\theta=\lambda$ directly to $a/\lambda=1/ …

Accepting a mismatched printed answer key value without re-deriving and checking it independently -- here, substituting $\theta=45°$ into $1/\sin\theta$ gives an unambiguous $\sqrt2\approx1.414$, which does not match the printed '1.27 …