In Young's double-slit experiment with a source of light of wavelength $6320 \ \mathring{A}$,the first maxima will occur when:

In Young's double-slit experiment,the value of $\lambda = 500\, nm$. The value of $d = 1\, mm$ and $D = 1\, m$. Then the minimum distance from the central maximum for which the intensity is half the maximum intensity will be:

Using Young's double slit experiment,a monochromatic light of wavelength $5000 \,\mathring A$ produces fringes of fringe width $0.5 \,mm$. If another monochromatic light of wavelength $6000 \,\mathring A$ is used and the separation between the slits is doubled,then the new fringe width will be ............... $mm$.

In Young's double slit experiment, the distance between the slits is $3 \text{ mm}$ and the slits are $2 \text{ m}$ away from the screen. Two interference patterns can be obtained on the screen due to light of wavelength $480 \text{ nm}$ and $600 \text{ nm}$ respectively. The separation on the screen between the $5^{th}$ order bright fringes of the two interference patterns is

In Young's double-slit experiment,the intensity of the central maximum is $I_0$. If one of the slits is closed,the intensity of the central maximum becomes .....

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: