In Young's double slit experiment,if the width (aperture) of the source slit $S$ is increased while keeping other parameters constant,then the interference fringes will:

In a double slit experiment,the distance between the slits is $0.1 \ cm$ and the screen is placed at $50 \ cm$ from the slits plane. When one slit is covered with a transparent sheet having thickness $t$ and refractive index $n = 1.5$,the central fringe shifts by $0.2 \ cm$. The value of $t$ is . . . . . . $cm$.

Two coherent point sources $S_1$ and $S_2$ vibrating in phase emit light of wavelength $\lambda$. The separation between them is $2 \lambda$ as shown in the figure. The first bright fringe is formed at $P$ due to interference on a screen placed at a distance $D$ from $S_1$ $(D >> \lambda)$. Find the distance $OP$.

Young's double slit experiment is performed with monochromatic light. $A$ thin film is introduced in front of one of the slits.

In a Young's double-slit experiment,light of wavelength $4800 \, \mathring A$ is used. If both slits are covered by transparent plates of the same thickness $t$ having refractive indices $\mu_1 = 1.5$ and $\mu_2 = 1.8$,and the central bright fringe shifts to the position of the fifth bright fringe,then the thickness of the plates is ....... $\mu m$.

In a Young's double-slit experiment,let $A$ and $B$ be the two slits. $A$ thin film of thickness $t$ and refractive index $\mu$ is placed in front of $A$. Let $\beta =$ fringe width. The central maximum will shift: