In which of the following is the interference due to the division of wave front?

In a Young's double-slit experiment,sodium light of wavelength $5890 \ \mathring A$ is used,and the angular width of the fringe is found to be $0.20^\circ$. If the angular width is to be increased by $10\%$,what is the required change in the wavelength?

In Young's double slit experiment,the distance between the sources is $1 \ mm$ and the distance between the screen and the source is $1 \ m$. If the fringe width on the screen is $0.06 \ cm$,then $\lambda = \dots \mathring{A}$.

In $\text{YDSE}$,$S_1$ and $S_2$ have the same intensity $I_0$. Column-$I$ shows the distance $x$ of a point $P$ from the central point $O$ on the screen,and Column-$II$ shows the intensity at $P$. Match Column-$I$ with Column-$II$. (Wavelength is $\lambda$)

Column-$I$	Column-$II$
$(A) x = \frac{D \lambda}{d}$	$(P) I_0$
$(B) x = \frac{D \lambda}{4d}$	$(Q) 2 I_0$
$(C) x = \frac{D \lambda}{3d}$	$(R) 3 I_0$
$(D) x = \frac{D \lambda}{6d}$	$(S) 4 I_0$

In Young's double slit experiment, the following figure shows that $Q$ is the position of the second bright fringe on the right side of point $O$. $P$ is the eleventh bright fringe on the other side measured from point $Q$. If the wavelength of light used is $6000 \text{ Å}$, then what will be the value of $S_1B$?

In Young's double slit experiment,the fringe width is $1 \times 10^{-4} \ m$. If the distance between the slit and screen is doubled,the distance between the two slits is reduced to half,and the wavelength is changed from $6.4 \times 10^{-7} \ m$ to $4.0 \times 10^{-7} \ m$,what will be the value of the new fringe width?