$A$ 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 $\text{ cm}$)?

The maximum number of possible interference maxima for slit separation equal to $1.8 \lambda$,where $\lambda$ is the wavelength of light used,in a Young's double slit experiment is

In a double slit interference experiment, the fringe width obtained with a light of wavelength $5900 \text{ Å}$ was $1.2 \text{ mm}$ for parallel narrow slits placed $2 \text{ mm}$ apart. In this arrangement, if the slit separation is increased by one-and-half times the previous value, then the fringe width is (in $ \text{ mm}$)

The source that illuminates the double-slit in a 'double-slit interference experiment' emits two distinct monochromatic waves of wavelengths $\lambda_1 = 500\,nm$ and $\lambda_2 = 600\,nm$. Each wavelength produces its own interference pattern on the screen. At the central point,where the path difference is zero,the maxima of both patterns coincide. As one moves away from the central region,the two fringe systems gradually go out of step. The combined fringe system becomes completely indistinct when a maximum of one wavelength coincides with a minimum of the other. This happens when the path difference in $nm$ is:

In $Y.D.S.E.$,how many maxima can be obtained on the screen if the wavelength of light used is $200 \ nm$ and $d = 700 \ nm$?

$A$ fringe width of a certain interference pattern is $\beta = 0.002 \text{ cm}$. What is the distance of the $5^{\text{th}}$ dark fringe from the center?