In Young's double slit experiment, the fringe width is $0.4 \text{ mm}$. What is the distance between $4^{th}$ dark band and $6^{th}$ bright band on the same side of the interference pattern (in $\text{ mm}$)?

In Young's double slit experiment,the fringe width is $1 \times 10^{-4} \ m$. If the distance between the slit and screen is doubled,the distance between the two slits is reduced to half,and the wavelength is changed from $6.4 \times 10^{-7} \ m$ to $4.0 \times 10^{-7} \ m$,what will be the value of the new fringe width?

If the Young's double slit experiment is performed with white light,then which of the following is not true?

$A$ beam of light consisting of wavelengths $650 \ nm$ and $550 \ nm$ illuminates the Young's double slits with separation of $2 \ mm$ such that the interference fringes are formed on a screen, placed at a distance of $1.2 \ m$ from the slits. The least distance of a point from the central maximum, where the bright fringes due to both the wavelengths coincide, is . . . . . . $\times 10^{-5} \ m$.

In Young's double slit experiment,the slits are $2\, mm$ apart and are illuminated by light of two wavelengths $\lambda_1 = 12000\, \mathring{A}$ and $\lambda_2 = 10000\, \mathring{A}$. At what minimum distance from the common central bright fringe on the screen $2\, m$ from the slits will a bright fringe from one interference pattern coincide with a bright fringe from the other? (in $mm$)

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: