In Young's double slit experiment,the $10^{th}$ order maximum is obtained at a point of observation in the interference pattern for a wavelength $\lambda_1 = 7000 \; \mathring{A}$. If the source is replaced by another one of wavelength $\lambda_2 = 5000 \; \mathring{A}$,what will be the order of the maximum at the same point (in $^{th}$)?

The fringe width in Young's double-slit experiment can be increased if we decrease

Monochromatic green light of wavelength $5 \times 10^{-7} \, m$ illuminates a pair of slits $1 \, mm$ apart. The separation of bright lines (fringe width) on the interference pattern formed on a screen $2 \, m$ away is ...... $mm$.

In a Young's double slit experiment, the separation between the two slits is $d$ and the wavelength of the light is $\lambda$. The intensity of light falling on slit $1$ is four times the intensity of light falling on slit $2$. Choose the correct choice(s).
$(A)$ If $d = \lambda$, the screen will contain only one maximum
$(B)$ If $\lambda < d < 2\lambda$, at least one more maximum (besides the central maximum) will be observed on the screen
$(C)$ If the intensity of light falling on slit $1$ is reduced so that it becomes equal to that of slit $2$, the intensities of the observed dark and bright fringes will increase
$(D)$ If the intensity of light falling on slit $2$ is increased so that it becomes equal to that of slit $1$, the intensities of the observed dark and bright fringes will increase

In Young's double slit experiment,the $8^{th}$ maximum with wavelength ${\lambda _1}$ is at a distance ${d_1}$ from the central maximum and the $6^{th}$ maximum with a wavelength ${\lambda _2}$ is at a distance ${d_2}.$ Then $({d_1}/{d_2})$ is equal to

In a Young's double slit experiment,the central point on the screen is