When two light waves of equal intensity superimpose, the maximum intensity obtained is $I$. If the intensity of one of the waves is quadrupled, then the maximum intensity obtained is

In Young's double slit experiment, if the distance between 5th bright and 7th dark fringes is 3 mm , then the distance between 5th dark and 7th bright fringes is

For an aperture of $5 \times 10^{-3} \mathrm{~m}$ and a monochromatic light of wavelength $\lambda$, the distance for which ray optics becomes a good approximation is 50 m , then $\lambda=$

In Young's double slit experiment with light of wavelength $\lambda$, the intensity of light at a point on the screen where the path difference becomes $\frac{\lambda}{3}$ is ( $I$ is intensity of the central bright fringe)