$A$ thin transparent plate of thickness $t$ and refractive index $\mu$ is introduced in the path of one of the two interfering waves. The path difference between these waves is changed by half the wavelength of light $\lambda$. The thickness of the plate is

In a Young's double-slit experiment,light of wavelength $4800 \, \mathring{A}$ is used. One slit is covered by a thin transparent plate of refractive index $1.4$ and the other slit is covered by another thin transparent plate of refractive index $1.7$. As a result,the central bright fringe shifts to the position where the fifth bright fringe was previously formed. If the thickness of both plates is the same,find the thickness in $\mu m$.

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 . . . . . . $cm$.

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 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 a Young's double-slit experiment,let $A$ and $B$ be the two slits. Thin films of thicknesses $t_A$ and $t_B$ and refractive indices $\mu_A$ and $\mu_B$ are placed in front of slits $A$ and $B$ respectively. If $\mu_A t_A = \mu_B t_B$,the central maximum will: