Discussion
Different spectral lines of the Balmer series (transitions \(n \rightarrow 2\), with \(n\) being the principal quantum number) fall one at a time on a Young’s double slit apparatus. The separation between the slits is \(d\) and the screen is placed at a constant distance from the slits. What factor should \(d\) be multiplied by to maintain a constant fringe width for various lines, as \(n\) takes different allowed values?
Answer: C
The factor is \(\frac{4n^{2}}{n^{2} - 4}\), up to a constant.
The fringe width is \(\beta = \lambda D/d\). With the screen distance fixed, keeping \(\beta\) constant requires \(d \propto \lambda\). For the Balmer series,
\[ \begin{aligned} \frac{1}{\lambda} &= R\left(\frac{1}{4} - \frac{1}{n^{2}}\right) = R\,\frac{n^{2} - 4}{4n^{2}} \\ \lambda &= \frac{1}{R} \cdot \frac{4n^{2}}{n^{2} - 4} \end{aligned} \]
so \(d\) must be scaled by \(4n^{2}/(n^{2} - 4)\). The option \((n^{2} - 4)/4n^{2}\) is proportional to \(1/\lambda\), the inverse of what is needed, and the forms with \(n^{2} + 4\) do not come from the Rydberg formula.
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