In Young's double-slit experiment,a thin mica sheet of thickness $12 \times 10^{-7} \ m$ is placed in the path of one of the interfering beams. It is observed that the central bright fringe shifts by a distance equal to the fringe width. If the wavelength of light used is $6 \times 10^{-7} \ m$,find the refractive index of the mica sheet.

In $Y.D.S.E.$,how many maxima can be obtained on the screen if the wavelength of light used is $200 \ nm$ and $d = 700 \ nm$?

In Young's double slit experiment,the ratio of intensity of maxima and minima in the interference experiment is $25 : 9$. The ratio of the widths of the two slits is:

In a Young's double slit experiment,the intensities at two points,for the path difference $\frac{\lambda}{4}$ and $\frac{\lambda}{3}$ ($\lambda$ being the wavelength of light used) are $I_1$ and $I_2$ respectively. If $I_0$ denotes the intensity produced by each one of the individual slits,then $\frac{I_1 + I_2}{I_0} = \dots$

The distance of the $5^{\text{th}}$ dark fringe from the center is $4 \, mm$. If $D = 2 \, m$ and $\lambda = 600 \, nm$,then the distance between the slits is (in $mm$):

In a Young's double slit experiment,slits are separated by $0.5 \ mm$,and the screen is placed $150 \ cm$ away. $A$ beam of light consisting of two wavelengths,$650 \ nm$ and $520 \ nm$,is used to obtain interference fringes on the screen. The least distance from the common central maximum to the point where the bright fringes due to both the wavelengths coincide is...... $mm$.