In Young's double slit experiment,the fringes are displaced by a distance $x$ when a glass plate of refractive index $1.5$ is introduced in the path of one of the beams. When this plate is replaced by another plate of the same thickness,the shift of fringes is $(3/2)x$. The refractive index of the second plate is

$A$ Young's double slit interference arrangement with slits $S_1$ and $S_2$ is immersed in water (refractive index $\mu_w = 4/3$) as shown in the figure. The positions of maxima on the surface of water are given by $x^2 = p^2 m^2 \lambda^2 - d^2$,where $\lambda$ is the wavelength of light in air (refractive index $\mu_a = 1$),$2d$ is the separation between the slits,and $m$ is an integer. The value of $p$ is

In a Young's double-slit experiment,the experiment is performed with blue light of wavelength $4360 \; \mathring{A}$ and green light of wavelength $5460 \; \mathring{A}$. If the distance of the $4^{th}$ bright fringe from the central maximum is $x$,then:

In an interference experiment,the $n^{\text{th}}$ bright fringe for light of wavelength $\lambda_1$ $(n=0, 1, 2, 3, \ldots)$ coincides with the $m^{\text{th}}$ dark fringe for light of wavelength $\lambda_2$ $(m=1, 2, 3, \ldots)$. The ratio $\frac{\lambda_1}{\lambda_2}$ is

When the distance between the slits is increased $4$ times in $YDSE$,the fringe width will

Interference fringes are observed on a screen by illuminating two thin slits $1 \, mm$ apart with a light source $(\lambda = 632.8 \, nm)$. The distance between the screen and the slits is $100 \, cm$. If a bright fringe is observed on a screen at a distance of $1.27 \, mm$ from the central bright fringe,then the path difference between the waves,which are reaching this point from the slits,is close to $.... \mu m$.