In Young's double slit experiment,if the wavelength of the light used is doubled and the distance between the two slits is halved,the resultant fringe width becomes $n$ times the initial fringe width. Find the value of $n$.

In a Young's double slit experiment,the path difference,at a certain point on the screen,between two interfering waves is $1/8^{th}$ of the wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to:

Which of the following statements is incorrect regarding interference fringes?

In Young's double slit experiment,light from two identical sources is superimposing on a screen. The path difference between the two light waves reaching a point on the screen is $\frac{7 \lambda}{4}$. The ratio of the intensity of the fringe at this point with respect to the maximum intensity of the fringe is:

In Young's double-slit experiment,$I_0$ is the intensity of the central maximum and $\beta$ is the fringe width. The intensity at a point $P$ at a distance $x$ from the center is:

Two slits separated by $4\, mm$ are illuminated by light of wavelength $6000\,\mathring{A}$. What will be the fringe width on a screen placed $2\, m$ away from the slits? (in $mm$)