Fraunhofer lines are obtained in

Mention the colours which have the least and the most wavelength in visible light.

Column $I$ shows four situations of standard Young's double slit arrangement with the screen placed far away from the slits $S_1$ and $S_2$. In each of these cases $S_1 P_0 = S_2 P_0$,$S_1 P_1 - S_2 P_1 = \lambda / 4$ and $S_1 P_2 - S_2 P_2 = \lambda / 3$,where $\lambda$ is the wavelength of the light used. In the cases $B, C$ and $D$,a transparent sheet of refractive index $\mu$ and thickness $t$ is pasted on slit $S_2$. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point $P$ on the screen from the two slits is denoted by $\delta(P)$ and the intensity by $I(P)$. Match each situation given in Column $I$ with the statement$(s)$ in Column $II$ valid for that situation.

Column $I$	Column $II$
$(A)$ No sheet	$(p)$ $\delta(P_0) = 0$
$(B)$ $(\mu-1)t = \lambda / 4$	$(q)$ $\delta(P_1) = 0$
$(C)$ $(\mu-1)t = \lambda / 2$	$(r)$ $I(P_1) = 0$
$(D)$ $(\mu-1)t = 3\lambda / 4$	$(s)$ $I(P_0) > I(P_1)$
	$(t)$ $I(P_2) > I(P_1)$

$A$ screen receives $3 \ W$ of radiant flux of wavelength $6000 \ \mathring{A}.$ One lumen is equivalent to $1.5 \times 10^{-3} \ W$ of monochromatic light of wavelength $5550 \ \mathring{A}.$ If the relative luminosity for $6000 \ \mathring{A}$ is $0.685$ while that for $5550 \ \mathring{A}$ is $1.00,$ then the luminous flux of the source is:

The solar spectrum during a complete solar eclipse is

Statement $(S)$: Using Huygen's eye-piece,measurements can be taken but are not correct.
Reason $(R)$: The cross wires,scale,and final image are not magnified proportionately because the image of the object is magnified by two lenses,whereas the cross-wire scale is magnified by one lens only.
Identify the correct option from the following: