Change in path difference for any point on screen is $$\left| {\mu - 1.8} \right|t.$$
For central maxima, phase difference $$= 0.$$
Hence,
$$\eqalign{ & d\sin \theta - \left| {\mu - 1.8} \right|t = 0 \cr & \Rightarrow d\theta = \left| {\mu - 1.8} \right|t\,\,\left[ {q\,{\text{is very small and is in radian}}} \right] \cr & \Rightarrow \left| {\mu - 1.8} \right| = \frac{{{{10}^{ - 3}} \times 0.1}}{{0.5 \times {{10}^{ - 3}}}} = 0.2 \cr & \Rightarrow \mu = 2\,\,{\text{or}}\,\,1.6 \cr} $$

For coherent sources :
$${I_1} = 4{I_0}$$
For incoherent sources
$$\eqalign{ & {I_2} = 2{I_0} \cr & \therefore \frac{{{I_1}}}{{{I_2}}} = \frac{2}{1} \cr} $$

There is air on both sides of the soap film.
∴ the reflections of the light produce a net $${180^ \circ }$$ phase shift.
The condition for bright fringes is $$2t = \left( {m + \frac{1}{2}} \right){\lambda _{{\text{film}}}}$$
$$\eqalign{ & t = \frac{{\left( {m + \frac{1}{2}} \right){\lambda _{{\text{film}}}}}}{2} = \frac{{\left( {m + \frac{1}{2}} \right)\lambda }}{{2n}} \cr & = \frac{{\left( {\frac{1}{2}} \right)\left( {650 \times {{10}^{ - 9}}m} \right)}}{{2\left( {1.41} \right)}} \cr & = 1.2 \times {10^{ - 7}}\,m \cr} $$

The reflections from the boundaries will cause a net $${180^ \circ }$$ phase shift.
The condition for bright fringes is $$2t = \left( {m + \frac{1}{2}} \right){\lambda _{{\text{film}}}}$$
Now, $$m= 124$$   since there is a bright fringe for $$m=0$$  and $${\lambda _{{\text{film}}}} = \frac{\lambda }{n}$$
$$\eqalign{ & t = \frac{{\left( {m + \frac{1}{2}} \right){\lambda _{{\text{film}}}}}}{2} = \frac{{\left( {m + \frac{1}{2}} \right)\lambda }}{{2n}} \cr & = \frac{{\left( {124 + \frac{1}{2}} \right)\left( {552 \times {{10}^{ - 9}}\,m} \right)}}{{2 \times \left( {1.00} \right)}} \cr & = 3.44 \times {10^{ - 5}}\,m \cr} $$

Path difference at $$P$$ is
$$\Delta x = 2\left( {\frac{x}{2}\cos \theta } \right) = x\cos \theta $$
For intensity to be maximum,
$$\eqalign{ & \Delta x = n\lambda \,\,\left( {n = 0,1,2,3,......} \right) \cr & {\text{or}}\,\,x\cos \theta = n\lambda \cr & {\text{or}}\,\,\cos \theta = \frac{{n\lambda }}{x} \geqslant 1 \cr & \therefore n \geqslant \frac{x}{\lambda } \cr & {\text{Subsituting }}x = 51,\,{\text{we get}} \cr & n \geqslant 5\,{\text{or}}\,\,n = 1,2,3,4,5,....... \cr} $$
Therefore in all four quadrants there can be 20 maxima. There are more maxima at $$\theta = {0^ \circ }\,{\text{and}}\,\theta = {180^ \circ }.$$    But $$n = 5$$  corresponds to $$\theta = {90^ \circ }\,{\text{and}}\,\theta = {270^ \circ }$$    which are coming only twice while we have multiplies it four times. Therefore, total number of maxima are still 20, i.e., $$n = 1$$  to 4 in four quadrants (total 16) plus more at $$\theta = {0^ \circ },{90^ \circ },{180^ \circ }\,{\text{and}}\,{270^ \circ }.$$

Constructive interference happens when $$2t = \left( {m - \frac{1}{2}} \right)\lambda .$$    The minimum value for $$m =$$  is $$m = 1,$$  the maximum value is the integer portion of
$$\eqalign{ & \frac{{2d}}{\lambda } + \frac{1}{2} = \frac{{2 \times 0.034 \times {{10}^{ - 3}}}}{{680 \times {{10}^{ - 9}}}} + \frac{1}{2} = 100.5 \cr & {m_{\max }} = 100 \cr} $$

$$\eqalign{ & {\beta _1} = \frac{{2.30}}{{20}} = \frac{{D{\lambda _1}}}{d} \cr & {\text{and}}\,\,{\beta _2} = \frac{{2.80}}{{30}} = \frac{{D{\lambda _2}}}{d} \cr & {\text{or}}\,\,\frac{{{\beta _1}}}{{{\beta _2}}} = \frac{{2.30 \times 30}}{{20 \times 2.80}} = \frac{{{\lambda _1}}}{{{\lambda _2}}} \cr & \therefore {\lambda _2} = 0.81\,{\lambda _1} = 0.81 \times 5890 = 4780\mathop {\text{A}}\limits^ \circ . \cr} $$

The intensity of light at any point of the screen where the phase difference due to light coming from the two slits is $$\phi $$ is given by
$$I = {I_0}{\cos ^2}\left( {\frac{\phi }{2}} \right)$$
where $${I_0}$$ is the maximum intensity.
NOTE : This formula is applicable when $${I_1} = {I_2}.$$   Here $$\phi = \frac{\pi }{3}$$
$$\eqalign{ & \therefore \,\,\frac{I}{{{I_0}}} = {\cos ^2}\frac{\pi }{6} = {\left( {\frac{{\sqrt 3 }}{2}} \right)^2} \cr & = \frac{3}{4} \cr} $$

Locus of equal path difference are lines running parallel to axis of the cylinder. Hence straight fringes will be observed.

The angle of incidence for total polarization is given by
$$\eqalign{ & \tan \theta = n \cr & \Rightarrow \,\,\theta = {\tan ^{ - 1}}n \cr} $$
Where $$n$$ is the refractive index of the glass.