In $YDSE$,the distance between slits is increased $10$ times,whereas their distance from the screen is halved. Then,the fringe width:

In a Young's double-slit experiment using a monochromatic source,what is the shape of the interference fringes formed on a screen?

The path difference between two interfering light waves meeting at a point on the screen is $\left(\frac{57}{2}\right) \lambda$. The band obtained at that point is

In the Young's double-slit experiment, the spacing between two slits is $0.1 \, mm$. If the screen is kept at a distance of $1.0 \, m$ from the slits and the wavelength of light is $5000 \, \mathring{A}$, then the fringe width is ........ $cm$.

Interference fringes are observed on a screen by illuminating two thin slits $1 \, mm$ apart with a light source $(\lambda = 632.8 \, nm)$. The distance between the screen and the slits is $100 \, cm$. If a bright fringe is observed on a screen at a distance of $1.27 \, mm$ from the central bright fringe,then the path difference between the waves,which are reaching this point from the slits,is close to $.... \mu m$.

In a Young's double slit experiment,the slit separation $d$ is $0.3 \text{ mm}$ and the screen distance $D$ is $1 \text{ m}$. $A$ parallel beam of light of wavelength $600 \text{ nm}$ is incident on the slits at angle $\alpha$ as shown in the figure. On the screen,the point $O$ is equidistant from the slits and the distance $PO$ is $11.0 \text{ mm}$. Which of the following statement$(s)$ is/are correct?