In a Young's double slit experiment, the slits are separated by $0.3 \, mm$ and the screen is $1.5 \, m$ away from the plane of slits. The distance between the fourth bright fringes on both sides of the central bright fringe is $2.4 \, cm$. The frequency of light used is $.......... \times 10^{14} \, Hz$.

In Young's double slit experiment,the light emitted from the source has $\lambda = 6.5 \times 10^{-7} \, m$ and the distance between the two slits is $1 \, mm$. The distance between the screen and the slits is $1 \, m$. The distance between the third dark fringe and the fifth bright fringe will be ......... $mm$.

If the source of light used in a Young's double slit experiment is changed from red to violet:

In a $YDSE$,light of two different wavelengths $(\lambda_1, \lambda_2)$ is incident normal to the plane of slits. The $n^{th}$ maxima of $\lambda_1$ coincides with the $m^{th}$ maxima of $\lambda_2$ exactly in front of one of the slits. Given $D = 1.5 \ m$,$d = 3 \ mm$,and $4500 \ \mathring{A} < \lambda_1, \lambda_2 < 7000 \ \mathring{A}$,then $n, m$ and $\lambda_1$ are:

In a Young's double-slit experiment,sources of equal intensities are used. The distance between the slits is $d$ and the wavelength of light used is $\lambda$ (where $\lambda \ll d$). Find the angular separation of the nearest points on either side of the central maximum where the intensity becomes half of the maximum value.

$A$ 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 . . . . . . $\times 10^{-5} \ m$.