In Young's double slit experiment using sodium light $(\lambda_1 = 5898 \ \text{\AA})$, $92$ fringes are seen. If a light of wavelength $\lambda_2 = 5461 \ \text{\AA}$ is used instead, how many fringes will be seen in the same field of view?

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 $X$?

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 $: 1$)

Two thin parallel slits are made in an opaque screen. When a monochromatic beam of light passes through them at normal incidence,the first bright fringe in the transmitted light occurs at $\pm 45^{\circ}$ with the original direction of the light beam on a distant screen when the apparatus is in air. When the apparatus is immersed in a liquid,the same bright fringe now occurs at $\pm 30^{\circ}$. The refractive index of the liquid is:

In a $YDSE$,if the slits are of unequal width:

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900 \text{ Å}$ was $1.2 \text{ mm}$ for parallel narrow slits placed $2 \text{ mm}$ apart. In this arrangement, if the slit separation is increased by one-and-half times the previous value, then the fringe width is (in $ \text{ mm}$)