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$\lambda_{1}$ and $\lambda_{2}$ are used to illuminate the slits. $\beta_{1}$ and $\beta_{2}$ are the corresponding fringe widths. The wavelength $\lambda_{1}$ can produce photoelectric effect when incident on a metal. But the wavelength $\lambda_{2}$ cannot produce photoelectric effect. The correct relation between $\beta_{1}$ and $\beta_{2}$ is
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Verified Answer
The correct answer is:
$\beta_{1} < \beta_{2}$
Fringe width $\beta \propto \lambda$. As given that $\lambda_{1}$ produces photo electric effect, so $\lambda_{1} < \lambda_{2}$.
Hence,
$\beta_{1} < \beta_{2}$
Hence,
$\beta_{1} < \beta_{2}$
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