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 relation between $\beta_1$ and $\beta_1$ is

• A. $\beta_1 < \beta_2$
• B. $\beta_1 = \beta_2$
• C. $\beta_1 > \beta_2$
• D. $\beta_1 \geq \beta_2$

Answer 1

Answer: (A)

$\beta_1 < \beta_2$

Answer 2

Fringe width $\beta \propto \lambda$. As given that $\lambda_{1}$ produces photo electric effect, so $\lambda_{1}