Question

Different spectral lines of the Balmer series (transitions $n \to 2$, 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?

• A. $\dfrac{n^2 - 4}{4n^2}$
• B. $\dfrac{n^2 + 4}{4n^2}$
• C. $\dfrac{4n^2}{n^2 - 4}$
• D. $\dfrac{4n^2}{n^2 + 4}$

Hint:

For maintaining a constant fringe width in a Young’s double slit experiment with varying wavelengths, the slit separation must be adjusted by a factor that compensates for the wavelength change with $n$.

Answer 1

The Correct Option is C

The fringe width in Young’s double slit experiment is given by: \[ \Delta y = \dfrac{\lambda L}{d}, \] where $\lambda$ is the wavelength, $L$ is the distance to the screen, and $d$ is the slit separation. The wavelength for a transition $n \to 2$ in the Balmer series is given by: \[ \lambda = \dfrac{R}{\left(\dfrac{1}{2^2} - \dfrac{1}{n^2}\right)}, \] where $R$ is the Rydberg constant. For the fringe width to be constant, the slit separation $d$ must be scaled by a factor that compensates for the change in $\lambda$ as $n$ changes. Thus, the required factor for scaling $d$ is $\dfrac{4n^2}{n^2 - 4}$, which corresponds to option (C).