In Young's double-slit interference experiment,the ratio of the slit widths is $1:25$. Find the ratio of the maximum to minimum intensity in the interference pattern.

In Young's double slit experiment,the angular width of fringes is $0.20^o$ for sodium light of wavelength $5890 \ \mathring A$. If the complete system is dipped in water,then the angular width of the fringes becomes....$^o$

In $YDSE$,let $S_1$ and $S_2$ be the two slits,and $C$ be the centre of the screen. If $\angle S_1 C S_2 = \theta$ and $\lambda$ is the wavelength,the fringe width 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 Young's double-slit experiment,using light of wavelength $\lambda = 5890 \ \mathring{A}$,the angular width of the fringe is $0.20^\circ$. To increase the angular width by $10\%$,the wavelength must be increased by .......... $\mathring{A}$.

$A$ beam of light consisting of two wavelengths $6500 \mathring{A}$ and $5200 \mathring{A}$ is used to obtain interference fringes in Young's double-slit experiment. The distance between slits is $2 \text{ mm}$ and the distance of the screen from the slits is $120 \text{ cm}$. What is the least distance from the central maximum where the bright fringes due to both wavelengths coincide (in $\text{ cm}$)?