An interference pattern was made by using red light. If the red light is replaced with blue light,the fringes will become

In Young's double slit experiment,the intensity at a point where the path difference is $\frac{\lambda}{6}$ is $I$. If $I_0$ denotes the maximum intensity,find the ratio $\frac{I}{I_0}$.

Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?

The fringe width in an interference pattern is $X$. The distance between the sixth dark fringe from one side of the central bright band to the fourth bright fringe on the other side is: (in $X$)

In a Young's double-slit experiment,the slits are separated by $0.28 \; mm$ and the screen is placed $1.4 \; m$ away. The distance between the central bright fringe and the fourth bright fringe is measured to be $1.2 \; cm$. Determine the wavelength of light used in the experiment (in $nm$).

In a Young's double-slit experiment,the distance between the two slits is $d = \lambda / 4$,where $\lambda$ is the wavelength of the light used. The initial phase difference is $\pi / 4$. What is the intensity at $\theta = 30^o$?