For dark fringe,
Path difference $$ = \frac{{\left( {2n - 1} \right)}}{2}\frac{\lambda }{2}$$

For $$i \approx {90^ \circ }$$  at air liquid interface we have by Snell’s law
$$\eqalign{ & \mu = \frac{{\sin {{90}^ \circ }}}{{\sin r}} \cr & \therefore \,\,\sin r = \frac{1}{\mu } \cr} $$
According to Brewster’s law, refractive index of liquid $$\left( \mu \right)$$  is equal to tangent of polarising angle
$$\eqalign{ & \because \,\,\tan {i_b} = \frac{{1.5}}{\mu } \cr & \therefore \,\,\sin {i_b} = \frac{{1.5}}{{\sqrt {{\mu ^2} + {{1.5}^2}} }} \cr} $$
Here, $$\sin r < \sin {i_b}$$

$$\eqalign{ & \therefore \,\,\frac{1}{\mu } \leqslant \frac{{1.5}}{{\sqrt {{\mu ^2} + {{\left( {1.5} \right)}^2}} }} \cr & {\text{or, }}\sqrt {{\mu ^2} + {{\left( {1.5} \right)}^2}} \leqslant 1.5 \times \mu \cr & \Rightarrow \,\,{\mu ^2} + {\left( {1.5} \right)^2} \leqslant {\left( {\mu \times 1.5} \right)^2} \cr & \Rightarrow \,\,\mu \geqslant \frac{3}{{\sqrt 5 }} \cr & {\text{i}}{\text{.e}}{\text{., minimum}} \cr} $$
value of $$\mu $$ should be $$\frac{3}{{\sqrt 5 }}$$

$$I = {I_0}{\cos ^2}\theta $$
Intensity of polarized light $$ = \frac{{{I_0}}}{2}$$
⇒ Intensity of untransmitted light $$ = {I_0} - \frac{{{I_0}}}{2} = \frac{{{I_0}}}{2}$$

A phase change of $$\pi $$ rad appears when the ray reflects at the glass-air interface. Also, the centre of the interference pattern is dark.

Equation of path difference form maxima in transmission (or weak reflection)
$$\eqalign{ & \Delta {P_{{\text{opt}}}} = 2{n_2}L = \frac{{{\lambda _{{\text{vacuum}}}}}}{2},\frac{{3{\lambda _{{\text{vacuum}}}}}}{2}\,...... \cr & \Rightarrow 2\left( {\frac{{{n_2}}}{{{n_1}}}} \right)L = \frac{\lambda }{2},\frac{{3\lambda }}{2},...... \cr & \Rightarrow L = \frac{\lambda }{{4{n_2}}}\,\,\left( {{\text{notice that }}\lambda = {\text{wavelength in medium is related to }}{\lambda _{{\text{vacuum}}}}\,{\text{as,}}\,{\lambda _{{\text{vacuum}}}} = {n_1}\lambda } \right) \cr} $$

$$I = {I_0}{\left( {\frac{{\sin \phi }}{\phi }} \right)^2}\,\,{\text{and }}\phi = \frac{\pi }{\lambda }\left( {b\sin \theta } \right)$$
When the slit width is doubled, the amplitude of the wave at the centre of the screen is doubled, so the intensity at the centre is increased by a factor 4.

The path difference at $$O,$$
$$\eqalign{ & \Delta x = d\sin \alpha = d\sin {30^ \circ } = \frac{{{{10}^{ - 3}}}}{2}m \cr & {\text{Now,}}\,\phi = \frac{{2\pi }}{\lambda }.\Delta x = \frac{{2\pi }}{{5000 \times {{10}^{ - 10}}}} \times \frac{{{{10}^{ - 3}}}}{2} = 2\pi \times {10^3} \cr & {\text{So}}\,I = {I_0} + {I_0} + 2\sqrt {{I_0}{I_0}} \cos \left( {2\pi \times {{10}^3}} \right) = 4{I_0} \cr} $$
The angular position of $$P,\tan \theta = \frac{1}{{\sqrt 3 }};$$    or $$\theta = {30^ \circ }.$$  It means path difference at $$P$$ is also $$\Delta x$$ and hence $$I = 4{I_0}$$

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

$$\eqalign{ & \therefore {X_0} = \frac{\beta }{\lambda }\left( {\mu - 1} \right)t \cr & \Rightarrow 5\beta = \frac{{\beta \left( {0.45} \right)t}}{{5890 \times {{10}^{ - 10}}}} \cr & \therefore t = \frac{{5 \times 5890 \times {{10}^{ - 10}}}}{{0.45}} = 6.544 \times {10^{ - 4}}\,cm \cr} $$