In a Young's double slit experiment,a laser light of $560\,nm$ produces an interference pattern with consecutive bright fringes' separation of $7.2\,mm$. Now another light is used to produce an interference pattern with consecutive bright fringes' separation of $8.1\,mm$. The wavelength of the second light is $......nm$.

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 a double-slit experiment, the angular width of a fringe for sodium light $(\lambda = 5890 \, \mathring{A})$ is $0.20^o$. If the fringe width increases by $10\%$, what is the change in the wavelength?

$A$ beam of electrons is used in $Y.D.S.E.$ The slit width is $d$. When the velocity of the electrons is increased,then:

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 $\mu_w = 4/3$)

Monochromatic green light of wavelength $5 \times 10^{-7} \, m$ illuminates a pair of slits $1 \, mm$ apart. The separation of bright lines (fringe width) on the interference pattern formed on a screen $2 \, m$ away is ...... $mm$.