Each plate reflects $25\%$ of the incident light intensity. When $AB$ and $A'B'$ are taken as the two slits in Young's double-slit experiment,what is the ratio of maximum to minimum intensity ${I_{\max }}/{I_{\min }}$ (in $: 1$)?

In a double-slit experiment,instead of taking slits of equal widths,one slit is made twice as wide as the other. Then,in the interference pattern:

In Young's double slit experiment, the aperture screen distance is $2 \, m$. The fringe width is $1 \, mm$. Light of $600 \, nm$ is used. If a thin plate of glass $(\mu = 1.5)$ of thickness $0.06 \, mm$ is placed over one of the slits, then there will be a lateral displacement of the fringes by $... \, cm$.

$A$ small transparent slab of refractive index $\mu = 1.5$ and thickness $L = d/4$ is placed along the path $AS_2$ (see figure). What will be the distance from $O$ of the principal maxima and the first minima on either side of the principal maxima obtained in the absence of the glass slab? Given $AC = CD = D$ and $S_1C = S_2C = d$ (where $d << D$).

The central fringe of an interference pattern produced by light of wavelength $6000 \, \mathring{A}$ is found to shift to the position of the fourth bright fringe after a glass plate of refractive index $1.5$ is introduced in front of one slit. The thickness of the glass plate would be ...... $\mu m$.

In a double slit experiment, when one of the slits is covered by a transparent mica sheet of refractive index $1.56$, the central fringe shifts to the position of $7^{th}$ bright fringe, obtained with both slits uncovered. If the light source wavelength is $450 \text{ nm}$, the thickness of mica sheet is $\alpha \times 10^{-9} \text{ m}$. The value of $\alpha$ is . . . . . . .