In Young's double slit experiment,the distance between the slits is reduced to half and the distance between the slit and the screen is doubled,then the fringe width:

In a Young's double-slit experiment,the light beam consists of two wavelengths $6500 \, \mathring{A}$ and $5200 \, \mathring{A}$. The distance between the slits is $2 \, mm$ and the distance between the plane of the slits and the screen is $120 \, cm$. What is the minimum distance from the central maximum where the bright fringes of both wavelengths coincide (in $, cm$)?

If the source of light used in a Young's Double Slit Experiment is changed from red to blue,then

In Young's double slit experiment,the seventh maximum with wavelength $\lambda_1$ is at a distance $d_1$ from the central maximum,and the same maximum with wavelength $\lambda_2$ is at a distance $d_2$. Then $d_1/d_2$ is:

In Young's double-slit experiment,the intensity at a point on the screen is $I_0/4$. Find the angular position of this point.

In Young's double-slit experiment,the slits are separated by $0.2 \, cm$ and illuminated by light of wavelength $\lambda = 5896 \, \mathring{A}$. What is the fringe width on a screen placed $1 \, m$ away from the plane of the slits? What will be the fringe width if the entire system is immersed in water (in $, mm$)? (Refractive index of water $\mu_w = 4/3$)