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 ..... $mm$.

In a Young's double slit experiment,the source $S$ and the two slits $A$ and $B$ are vertical,with slit $A$ above slit $B$. The fringes are observed on a vertical screen $K$. The optical path length from $S$ to $B$ is increased very slightly (by introducing a transparent material of higher refractive index) and the optical path length from $S$ to $A$ is not changed. As a result,the fringe system on $K$ moves:

In Young's double-slit experiment,for a distance of $1.8 \lambda$ between the two slits,what is the maximum number of possible interference maxima? Here,$\lambda$ is the wavelength of light.

Two wavelengths $\lambda_1 = 450 \ nm$ and $\lambda_2 = 650 \ nm$ are used in Young's double slit experiment. The minimum order of fringe produced by $\lambda_2$ which overlaps with a fringe produced by $\lambda_1$ is $n$. The value of $n$ is . . . . . . .

What happens to the fringe width in Young's double-slit experiment when it is performed in water instead of air?

Young's double slit experiment is carried out by using green,red,and blue light,one color at a time. The fringe widths recorded are $\beta_G, \beta_R$,and $\beta_B$,respectively. Then: