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

When monochromatic light of wavelength $600 nm$ is used in Young's double slit experiment,the fifth order bright fringe is formed at $6 mm$ from the central bright fringe on the screen. If the experiment is conducted with light of wavelength $400 nm$,the third order bright fringe will be located at: (in $mm$)

$A$ fringe width of a certain interference pattern is $\beta = 0.002 \text{ cm}$. What is the distance of the $5^{\text{th}}$ dark fringe from the center?

$A$ double slit experiment is immersed in water of refractive index $1.33$. The slit separation is $1 \,mm$, and the distance between the slit and the screen is $1.33 \,m$. The slits are illuminated by light of wavelength $6300 \,Å$. The fringe width is:

In Young's double-slit experiment,the intensity of light at a point on the screen where the path difference is $\lambda$ is $k$ units; $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\lambda/4$ will be:

The ratio of intensities at two points $P$ and $Q$ on the screen in a Young's double-slit experiment,where the phase differences between two waves of the same amplitude are $\pi / 3$ and $\pi / 2$ respectively,is: