$$n$$^th minimum has a distance from the centre $$ = x = \left( {2n + 1} \right)\frac{1}{2}\frac{{\lambda D}}{d}$$
For a point on the screen directly in front of one of the slits, $$x = \frac{d}{2}$$
$$\therefore $$ for minimum intensity in front of one of the slits $$ = \frac{d}{2} = \left( {2n + 1} \right)\frac{\lambda }{2}\frac{D}{d}$$
$$\therefore \lambda = \frac{{{d^2}}}{{\left( {2n + 1} \right)D}}$$

At the area of total darkness minima will occur for both the wavelengths.
$$\eqalign{ & \therefore \frac{{\left( {2n + 1} \right)}}{2}{\lambda _1} = \frac{{\left( {2m + 1} \right)}}{2}{\lambda _2} \cr & \Rightarrow \left( {2n + 1} \right){\lambda _1} = \left( {2m + 1} \right){\lambda _2} \cr & {\text{or}}\,\,\frac{{\left( {2n + 1} \right)}}{{\left( {2m + 1} \right)}} = \frac{{560}}{{400}} = \frac{7}{5} \cr & {\text{or}}\,\,10n = 14m + 2 \cr} $$
by inspection for $$m= 2,n =3$$   and for $$m=7,n= 10,$$   the distance between them will be the distance between such points.
i.e., $$\Delta s = \frac{{D{\lambda _1}}}{d}\left\{ {\frac{{\left( {2{n_2} + 1} \right) - \left( {2{n_1} + 1} \right)}}{2}} \right\}$$
put $${n_2} = 10,{n_1} = 3$$
On solving we get, $$\Delta s = 28\,\,mm.$$

For common maxima, $${n_1}{\lambda _1} = {n_2}{\lambda _2}$$
$$\eqalign{ & \Rightarrow \frac{{{n_1}}}{{{n_2}}} = \frac{{{\lambda _2}}}{{{\lambda _1}}} = \frac{{520 \times {{10}^{ - 9}}}}{{650 \times {{10}^{ - 9}}}} = \frac{4}{5} \cr & {\text{For}}\,\,{\lambda _1} \cr & y = \frac{{{n_1}{\lambda _1}D}}{d},{\lambda _1} = 650\,nm \cr & y = \frac{{4 \times 650 \times {{10}^{ - 9}} \times 1.5}}{{0.5 \times {{10}^{ - 3}}}} \cr & {\text{or,}}\,{\text{ }}y = 7.8\,mn \cr} $$

The electron beam will be diffracted and the maxima is obtained at $$y = 0.$$  Also the distance between the first minima on both side will be greater than $$d.$$

when slits are of equal width.
$$\eqalign{ & {I_{\max }}\, \propto \,{\left( {a + a} \right)^2}\left( { = 4{a^2}} \right) \cr & {I_{\min }}\, \propto \,{\left( {a - a} \right)^2}\left( { = 0} \right) \cr} $$
When one slit's width is twice that of other
$$\eqalign{ & \frac{{{I_1}}}{{{I_2}}} = \frac{{{W_1}}}{{{W_2}}} = \frac{{{a^2}}}{{{b^2}}} \cr & \Rightarrow \,\,\frac{W}{{2W}} = \frac{{{a^2}}}{{{b^2}}} \cr & \Rightarrow \,\,b = \sqrt 2 a \cr & \therefore \,\,{I_{\max }}\, \propto \,{\left( {a + \sqrt 2 a} \right)^2} = \left( {5.8{a^2}} \right) \cr & {I_{\min }}\, \propto \,{\left( {\sqrt 2 a - a} \right)^2} = \left( {0.17{a^2}} \right) \cr} $$

For maxima $$d\sin \theta = n\lambda $$
$$\eqalign{ & \sin \theta = \frac{{n\lambda }}{d} = \frac{{8\lambda }}{{10\lambda }} \cr & \Rightarrow \sin \theta = \frac{4}{5} \cr & \Rightarrow \tan \theta = \frac{4}{3} \cr & {\text{Also}}\,\,\tan \theta = \frac{y}{D} \cr & \therefore y = \frac{{4D}}{3} \cr} $$

Fringe width $$\omega = \frac{{\lambda D}}{d}.$$
When the apparatus is immersed in a liquid, then $$\lambda $$ will decrease $$\mu $$ times and hence $$\omega $$ is reduced $$\mu $$ (refractive index) times. $$10\omega ' = \left( {5.5} \right)\omega \,\,{\text{or}}\,\,10\lambda '\left( {\frac{D}{d}} \right) = \left( {5.5} \right)\frac{{\lambda D}}{d}\,\,{\text{or}}\,{\kern 1pt} \frac{\lambda }{{\lambda '}} = \frac{{10}}{{5.5}} = \mu $$
$${\text{or}}\,\,\mu = 1.8$$

$$\eqalign{ & {\text{Intensity}} \propto {\left( {{\text{amplitude}}} \right)^2} \cr & \frac{{{I_1}}}{{{I_2}}} = \frac{{16}}{9} = \frac{{a_1^2}}{{a_2^2}} \cr & \Rightarrow {a_1} = 4;{a_2} = 3 \cr} $$
Therefore the ratio of intensities of bright and dark parts
$$\frac{{{I_{{\text{Bright}}}}}}{{{I_{{\text{Dark}}}}}} = \frac{{{{\left( {{a_1} + {a_2}} \right)}^2}}}{{{{\left( {{a_1} - {a_2}} \right)}^2}}} = \frac{{{{\left( {4 + 3} \right)}^2}}}{{{{\left( {4 - 3} \right)}^2}}} = \frac{{49}}{1}$$

In the medium, the refractive index will decrease from the axis towards the periphery of the beam.
Therefore, the beam will move as one move from the axis to the periphery and hence the beam will converge.

According to malus law, intensity of emerging beam is given by,
$$\eqalign{ & I = {I_0}{\cos ^2}\theta \cr & {\text{Now, }}{I_{A'}} = {I_A}{\cos ^2}{30^ \circ } \cr & {I_{B'}} = {I_B}{\cos ^2}{60^ \circ } \cr & {\text{As, }}{I_{A'}} = {I_{B'}} \cr & \Rightarrow \,\,{I_A} \times \frac{3}{4} = {I_B} \times \frac{1}{4} \cr & \therefore \,\,\frac{{{I_A}}}{{{I_B}}} = \frac{1}{3} \cr} $$