In Young's double slit interference experiment,the slit separation is made $3$ fold. The fringe width becomes

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 $YDSE$ experiment,the $4^{\text{th}}$ dark band is formed opposite to one of the slits. The wavelength of light used is ($d=$ distance between the slits,$D=$ distance between source and the screen)

Two slits are made $10 \ mm$ apart and the screen is placed $1.5 \ m$ away. What is the fringe separation when a wavelength of $7000 \ \mathring{A}$ is used (in $\mu m$)?

In Young's double slit experiment,carried out with light of wavelength $5000 \ \mathring{A}$,the distance between the slits is $0.3 \ \text{mm}$ and the screen is at $200 \ \text{cm}$ from the slits. The central maximum is at $x=0 \ \text{cm}$. The value of $x$ for the third maxima is ............. $\text{mm}$.

White light is passed through a double slit and interference is observed on a screen $1.5 \, m$ away. The separation between the slits is $0.3 \, mm$. The first violet and red fringes are formed $2.0 \, mm$ and $3.5 \, mm$ away from the central white fringes. The difference in wavelengths of red and violet light is $.... \, nm$.