In Young's double slit experiment, the wavelength of light used is $\lambda$. The intensity on the screen at a point for path difference '$\lambda$' is 'X'. The intensity at the point for path difference $\left(\dfrac{\lambda}{6}\right)$ is ($\cos 180^\circ = -1$, $\cos 30^\circ = \dfrac{\sqrt{3}}{2}$)

In a single slit diffraction pattern, the distance between the plane of the slit and the screen is $1.4$ m. The width of the slit is $0.66$ mm. The second maximum is formed at the distance of $2.8$ mm from the center of the screen. The wavelength of light used is

In Young's double slit experiment, for the $n^{\text{th}}$ dark fringe ($n = 1, 2, 3, \ldots$) the phase difference of the interfering waves in radian will be

A photoelectric surface is illuminated successively by monochromatic light of wavelength $\lambda$ and $(\lambda/3)$. If the maximum kinetic energy of the emitted photo electrons in the second case is $4$ times that in the first case, the work function of the surface of the material is ($h$ = Planck's constant, $c$ = speed of light)