In Young's double slit experiment,the amplitudes of two sources are $3a$ and $a$ respectively. The ratio of intensities of bright and dark fringes will be (in $:1$)

The angular width of a fringe in a double slit experiment is found to be $0.2^{\circ}$ on a screen $1 \ m$ away. The wavelength of light used is $600 \ nm$. What is the change in the angular width of the fringe if the entire measurement system is immersed in water (in $^{\circ}$)? (Use refractive index of water as $\mu = \frac{4}{3}$)

In a Young's double slit experiment,the fringe width is found to be $0.4 \, mm$. If the whole apparatus is immersed in water of refractive index $4/3$ without disturbing the geometrical arrangement,the new fringe width will be

In Young's double-slit experiment,the intensity at a point is $(1/4)$ of the maximum intensity. The angular position of this point is:

The figure shows a double slit experiment where $P$ and $Q$ are the slits. The path lengths $PX$ and $QX$ are $n\lambda$ and $(n + 2)\lambda$ respectively,where $n$ is a whole number and $\lambda$ is the wavelength. Taking the central fringe as zero,what is formed at $X$?

Two coherent sources $O_1$ and $O_2$ in Young's double slit experiment are illuminated with monochromatic light of wavelength $5000 \text{ Å}$. If a second order dark fringe is formed at a point $R$ on the screen,the path difference $O_1 R \sim O_2 R$ is (in $\mu m$)