In Young's double-slit experiment,the fringe width is $12 \ mm$. If the entire arrangement is placed in water of refractive index $\frac{4}{3}$,then the fringe width becomes (in $mm$):

In a Young's double-slit experiment,the fringe width is $0.8 \, mm$ when light of wavelength $6000 \, \mathring{A}$ is used,and the screen is at a distance of $2.5 \, m$. If the experiment is performed in a liquid with a refractive index of $1.6$,the new fringe width will be.....$mm$.

In Young's double slit experiment, the width of the second slit is double the width of the first slit. Consequently, the amplitude of the light from the two slits is different. If $I_m$ is the maximum intensity, the resultant intensity $I$ when they interfere with a phase difference of $\phi$ is given by:

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,if the distance between the slits is twice the wavelength,the number of possible interference maxima is:

In a Young's double slit experiment,the slit separation is $0.2 \ cm$,the distance between the screen and slit is $1 \ m$. The wavelength of the light used is $5000 \ \mathring{A}$. The distance between two consecutive dark fringes (in $mm$) is