Question

${\lambda }_1$ 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 answer relation between ${\beta }_1$ and ${\beta }_2$ is.

• A. ${\beta }_1<{\beta }_2$
• B. ${\beta }_1={\beta }_2$
• C. ${\beta }_1>{\beta }_2$
• D. ${\beta }_1\ge {\beta }_2$

Answer 1

The correct option is A ${\beta }_1<{\beta }_2$

Fringe width, $\beta =\frac{\lambda D}{d}$ $⟹\beta \propto \lambda$

As wavelength ${\lambda }_2$ cannot produce photoelectric effect but ${\lambda }_1$ does.

$⟹$ ${\lambda }_1<{\lambda }_2$

$⟹$ ${\beta }_1<{\beta }_2$