Given, focal length of objective, f₀ = 8mm = 0.8 cm
Focal length of eyepiece, f_e = 2.5 cm
Distance of object from objective, u₀ = - 9 mm = - 0.9 cm
Distance of image from eyepiece, v_e = - D = - 25 cm
Using lens formula for eyepiece, \(\frac { 1 }{ { v }_{ e } } -\frac { 1 }{ { u }_{ e } } =\frac { 1 }{ { f }_{ e } } \\ \) 
or \(\frac { 1 }{ { u }_{ e } } =\frac { 1 }{ { v }_{ e } } -\frac { 1 }{ { f }_{ e } } =-\frac { 1 }{ 25 } -\frac { 1 }{ 2.5 } =\frac { -1-10 }{ 25 } =-\frac { 11 }{ 25 } \\ \) 
u_e = - 2.27 cm
Using lens formula for objective, \(\frac { 1 }{ { v }_{ e } } -\frac { 1 }{ { u }_{ 0 } } =\frac { 1 }{ { f }_{ 0 } } \)
or \(\frac { 1 }{ { v }_{ 0 } } =\frac { 1 }{ { f }_{ 0 } } +\frac { 1 }{ { u }_{ 0 } } =-\frac { 1 }{ 0.8 } -\frac { 1 }{ 0.9 } =-\frac { 0.9-0.8 }{ 0.72 } =\frac { 0.1 }{ 0.72 } \\ \)  
Distance of image for objective lens, v₀ = 7.2 cm
Separation between two lenses,
\(L=|{ u }_{ e }|+|{ v }_{ 0 }|=2.27+7.2=9.47 \ cm\) 
Magnifying power of co,pound microscope,
\(m=\frac { { v }_{ o } }{ { u }_{ 0 } } (1+\frac { D }{ { f }_{ e } } )=\frac { 7.2 }{ 0.9 } (1+\frac { 25 }{ 2.5 } )=88\)

Intensity can be found out, if we know the phase difference. Phase difference can be calculated with the help of path difference. So first of all, path difference will be calculated.
Given, \(OP={ y }_{ n }\)
The distance OP equals one-third of frings width of the pattern.
i.e  \({ y }_{ n }=\frac { \beta }{ 3 } =\frac { 1 }{ 3 } \left( \frac { D\lambda }{ d } \right) =\frac { D\lambda }{ 3d } \)
\(\Rightarrow \ \frac { { dy }_{ n } }{ D } =\frac { \lambda }{ 3 } \)
Path difference=\({ S }_{ 2 }P-{ S }_{ 1 }P=\frac { { dy }_{ n } }{ D } =\frac { \lambda }{ 3 } \)
\(\therefore \)  Phase difference, \(\phi =\frac { 2\pi }{ \lambda } \times \)path difference
\(\frac { 2\pi }{ \lambda } \times \frac { \lambda }{ 3 } =\frac { 2\pi }{ 3 } \)
If intensuty at central frings is \(I^o\), then intensity at a point P, where phase difference \(\phi \) is given by
\(I=I^o\)\({ cos }^{ 2 }\phi \)
\(I=I^o\)\({ \left( cos\frac { 2\pi }{ 3 } \right) }^{ 2 }\)
= \(I^o\)\({ \left( -cos\frac { \pi }{ 3 } \right) }^{ 2 }\)=\(I^o\)\({ \left( -\frac { 1 }{ 2 } \right) }^{ 2 }\)=\(\quad \frac { { I }_{ \circ } }{ 4 } \)
Hence, the intensity at point p would be \( \frac { { I }_{ \circ } }{ 4 } \).

-18.3 cm

When angle of incidence is equal to Brewster's angle, the transmitted light is unpolarised and reflected light is plane polarised. Consider the diagram in which unpolarised light is represented by dot and plane polarised light is represented by arrows.

Polarised by reflection occurs, when the angle of incidence is equal to the Brewster's angle. i.e. \(tani_{ B }=^{ 1 }{ \mu }_{ 2 }=\frac { { \mu }_{ 2 } }{ { \mu }_{ 1 } } ,\ where{ \ \mu }_{ 2 }<{ \mu }_{ 1 }\)
When the light rays travel in such a medium, the critical angle is expressed as \(sini_{ c }=\frac { { \mu }_{ 2 } }{ { \mu }_{ 1 } } \ where,{ \ \mu }_{ 2 }<{ \mu }_{ 1 }\)
As, \(|tani_{ B }|>|sini_{ c }|\) for large angles 
Thus, definitely polarisation by reflection will occur.

As we know that,  \(I={ I }_{ av }\pm 0.05{ I }_{ av }\)
\(\Rightarrow \)   \({ I }_{ max }={ I }_{ av }\left( 1+0.05 \right) =1.05{ I }_{ av }\)    
 \({ I }_{ min }={ I }_{ av }\left( 1-0.05 \right) =0.95{ I }_{ av }\) 
Dividing Eqs. (i) and (ii), we get
\(\Rightarrow \)  \(\frac { { I }_{ max } }{ { I }_{ min } } =\frac { 1.05 }{ 0.95 } \) 
\(\Rightarrow \) \(\left( \frac { r+1 }{ r-1 } \right) ^{ 2 }=\frac { 105 }{ 95 } \) 
\(\Rightarrow \)     \({ r }^{ 2 }+1+2r)95=105({ r }^{ 2 }+1-2r)\) 
\(\Rightarrow \)   \({ 10r }^{ 2 }-400r+10=0\Rightarrow r=40\) 
\(\frac { { I }_{ 1 } }{ { I }_{ 2 } } ={ r }^{ 2 }=({ 40) }^{ 2 }=1600:1\)