In a Young's double slit experiment,the width of one of the slits is three times the other slit. The amplitude of the light coming from a slit is proportional to the slit width. Find the ratio of the maximum to the minimum intensity in the interference pattern.

In a Young's double-slit experiment,the slits are placed $0.320 \, mm$ apart. Light of wavelength $\lambda = 500 \, nm$ is incident on the slits. The total number of bright fringes that are observed in the angular range $-30^\circ \le \theta \le 30^\circ$ is:

In Young's double-slit experiment,the two slits are $2 \, mm$ apart and are illuminated by a mixture 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 will a bright fringe from one interference pattern coincide with a bright fringe from the other? The distance between the slits and the screen is $2 \, m$.

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:

In a $Young's$ double-slit experiment,light of wavelength $6000 \, \mathring{A}$ is used. If the third dark fringe is formed at point $P$ on the screen,then the path difference $(S_1P - S_2P)$ is ......... $\mu m$.

In Young's double-slit experiment, light of wavelength $4000 \, Å$ is used to produce bright fringes of width $0.6 \, mm$ at a distance of $2 \, m$ from the slits. If the whole apparatus is immersed in a liquid of refractive index $1.5$, what will be the new fringe width in $mm$?