In a Young's double slit experiment, the intensities at two points, for the path difference $\dfrac{\lambda}{4}$ and $\dfrac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits, then $\dfrac{I_1 + I_2}{I_0} = $
$\left(\cos 45^\circ = \dfrac{1}{\sqrt{2}}, \cos 60 = \dfrac{1}{2}\right)$

Two light waves amplitudes in the ratio $3:1$ produce interference. The ratio of the maximum to minimum intensity is

Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light rays ($v_1 > v_2$). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of $k : 1$, then what is the threshold frequency of the metallic surface?

A photon and an electron have equal energy E. The ratio of $\lambda$(electron) to $\lambda$(photon) is proportional to