In Young's double slit experiment,the width of fringes can be increased if we decrease the

$A$ 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 Young's double-slit experiment,the distance between the two slits is $2 \times 10^{-3} \, m$ and the distance between the slits and the screen is $2.5 \, m$. The wavelength of the light used ranges from $2000 \, \mathring{A}$ to $9000 \, \mathring{A}$. What wavelength (in $\mathring{A}$) will form a bright fringe at a distance of $10^{-3} \, m$ from the central maximum?

In an interference pattern,the $(n + 4)^{th}$ order blue bright fringe and the $n^{th}$ order red bright fringe coincide at a point. If the wavelengths of red and blue light are $7800 \, \mathring{A}$ and $5200 \, \mathring{A}$ respectively,then the value of $n$ is . . . . . .

In $YDSE$, the distance of the slits from the screen is increased by $25 \%$ and the separation between the slits is halved. If $W$ represents the original fringe width, the new fringe width is (in $\,W$)

$A$ Young's double-slit experimental setup is immersed in water of refractive index $1.33$. It has a slit separation of $1 \ mm$ and the distance between the slits and the screen is $1.33 \ m$. If the wavelength of incident light on the slits is $6300 \ \mathring{A}$,then the fringe width on the screen is: