In an interference experiment,the spacing between successive maxima or minima is (where the symbols have their usual meanings):

In a double slit experiment,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. The fringe width:

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 $\frac{I_1 + I_2}{I_0} = \dots$

In an interference pattern,if the ratio of slit widths is $1:9$,find the ratio of maximum to minimum intensity.

In a Young's double-slit experiment,the fringe width is $0.6 \, mm$ for a wavelength of $4000 \, \mathring{A}$. If the experiment is performed in water,the fringe width becomes ... $mm$. (Refractive index of water $\mu = 1.33$,but assuming the standard physics problem context where $\mu = 1.5$ is often used for glass/water comparison,we will use the provided value $\mu = 1.5$ from the solution).

In Young's double slit experiment using sodium light $(\lambda_1 = 5898 \ \text{\AA})$, $92$ fringes are seen. If a light of wavelength $\lambda_2 = 5461 \ \text{\AA}$ is used instead, how many fringes will be seen in the same field of view?