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 $\mu_w = 4/3$)

$A$ double slit experiment is performed by using light of wavelength $6000 \, Å$. If the distance of the screen is $1 \, m$ and the slits are $0.1 \, cm$ apart, calculate the angular position of the $10^{th}$ bright fringe.

In Young's double slit experiment,the phase difference between the light waves reaching the third bright fringe from the central fringe is: $(\lambda = 6000 \ \mathring{A})$

In a Young's double-slit experiment,the fringe width is $0.6 \, mm$ for a wavelength of $4000 \, \mathring{A}$. If the experiment is performed in water,the fringe width becomes ... $mm$. (Refractive index of water $\mu = 1.33$,but assuming the standard physics problem context where $\mu = 1.5$ is often used for glass/water comparison,we will use the provided value $\mu = 1.5$ from the solution).

In Young's double-slit experiment,if a metal plate of thickness $t$ is placed in the path of one of the rays,then:

In a Young's double-slit experiment,the fringe width is $0.2 \, mm$. If the wavelength of light is increased by $10\%$ and the distance between the slits is increased by $10\%$,then the new fringe width will be ..... $mm$.