In a Young's double-slit experiment,$D$ is the distance of the screen from the slits and $d$ is the separation between the slits. The distance of the nearest point to the central maximum where the intensity is the same as that due to a single slit is equal to:

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 $.... \, nm$.

In a Young's double-slit experiment,the slit separation is $0.589 \, mm$ using sodium light. What is the angular separation between the third maximum and the central maximum? $(\lambda = 589 \, nm)$

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.

In a double slit experiment with monochromatic light,fringes are obtained on a screen placed at some distance from the plane of slits. If the screen is moved by $5 \times 10^{-2} \,m$ towards the slits,the change in fringe width is $3 \times 10^{-3} \,cm$. If the distance between the slits is $1 \,mm$,then the wavelength of the light will be . . . . . . $nm$.

In a Young's experiment,two coherent sources are placed $0.90 \, mm$ apart and the fringes are observed one metre away. If it produces the second dark fringe at a distance of $1 \, mm$ from the central fringe,the wavelength of monochromatic light used would be