If one of the slits of a standard $YDSE$ apparatus is covered by a thin parallel-sided glass slab so that it transmits only one-half of the light intensity of the other, then:

The fringe widths are found to be $\omega_1$ and $\omega_2$ respectively if a Young's double slit experiment is performed in media of refractive indices $n_1$ and $n_2$ respectively. The correct statement is:

The figure shows a double-slit experiment. Each slit has a width $W$. $A$ thin glass slab of thickness $\delta$ and refractive index $\mu$ is placed in front of one of the slits. The intensity at the central point $C$ is measured as a function of the thickness $\delta$. At what thickness $\delta$ will the intensity at $C$ be minimum?

Consider the figure (not drawn to scale) in which a converging lens of radius $R = 1 \ cm$ and focal length $f = 20 \ cm$ is cut in the middle. The upper part is lifted up by $d = 1 \ mm$ and the lower part is pulled down by the same distance. The gap between them is blocked by an opaque sheet. $A$ point light source with wavelength $\lambda = 500 \ nm$ is placed on the optical axis at a distance of $2f$ from the split lens. $A$ large screen is placed at $L = 1 \ m$ from the right focus of the lens. Find the approximate number of interference fringes on the screen.

The central fringe of the interference pattern produced by light of wavelength $6000 \mathring A$ is found to shift to the position of the $4^{\text {th}}$ bright fringe after a glass plate of refractive index $1.5$ is introduced in front of one slit in Young's experiment. The thickness of the glass plate will be ......... $\mu m$.

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