In a Young's double slit experiment,the ratio of the slit widths is $4 : 1$. The ratio of the intensity of maxima to minima,close to the central fringe on the screen,will be

$A$ 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 an interference experiment,the spacing between successive maxima or minima is (where the symbols have their usual meanings):

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:

In Young's double-slit interference experiment,the distance between two sources is $0.1 \ mm$. The distance of the screen from the sources is $20 \ cm$. The wavelength of light used is $5460 \ \mathring{A}$. What is the angular position of the first dark fringe in degrees?

In $YDSE$,the angular width of a fringe formed on a distant screen is $1^o$. The slit separation is $0.01\,mm$. The wavelength of light is: