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
\(\Rightarrow \)  \(\frac { { I }_{ max } }{ { I }_{ min } } =\frac { 1.05 }{ 0.95 } \) 
\(\Rightarrow \) \(\left( \frac { r+1 }{ r-1 } \right) ^{ 2 }=\frac { 105 }{ 95 } \) 
\(\Rightarrow \)     \({ r }^{ 2 }+1+2r)95=105({ r }^{ 2 }+1-2r)\) 
\(\Rightarrow \)   \({ 10r }^{ 2 }-400r+10=0\Rightarrow r=40\) 
\(\frac { { I }_{ 1 } }{ { I }_{ 2 } } ={ r }^{ 2 }=({ 40) }^{ 2 }=1600:1\)

In normal adjustments, m = \(|\frac { { f }_{ 0 } }{ { f }_{ e } } |=20\\ \)
Also, length of telescope, f₀ + f_e = 105
20f₀ + f _e = 105
21f _e = 105
f _e = 5 cm
f₀ = 20 f_e = 20 x 5 = 100 cm

(i) By Malus law, intensity of emergent light from \({ P }_{ 2 }\) is \(I={ I }_{ \circ }cos^{ 2 }\theta \) , where \(\theta \) is the angle between \({ P }_{ 1 }and{ P }_{ 2 }\)|
When  \(\theta ={ 90 }^{ \circ }\) 
\(\\ I={ I }_{ \circ }\times 0\)
Intensity of emergent light, I = 0
(ii) Intensity of light from \({ P }_{ 3 }\)
 \(=\left( \frac { { I }_{ \circ } }{ 2 } cos^{ 2 }\beta \right) \left[ cos^{ 2 }\left( { 90 }^{ \circ }-\beta \right) \right] \)

\(=\frac { { I }_{ \circ } }{ 2 } cos^{ 2 }\beta sin^{ 2 }\beta =\frac { { I }_{ \circ } }{ 8 } sin^{ 2 }2\beta \)
\(As,\ \frac { { I }_{ \circ } }{ 8 } sin^{ 2 }2\beta =\frac { { I }_{ \circ } }{ 8 } (given)\)
\(so, \ \left( sin2\beta \right) ^{ 2 }=1 \ \Rightarrow 2\beta ={ 90 }^{ \circ } \ \Rightarrow \beta ={ 45 }^{ \circ }\)

Any ray entering at an angle i shall be guided along AC, if the ray makes an angle \(\phi \) with the face AC greater than the critical angle as the principle of total internal reflection,\(\phi \)+r = 90^o, therefore, sin \(\phi \) = cos r.
\(\Rightarrow \ sin\phi \ge \frac { 1 }{ \mu } \ \Rightarrow \ cos\ r\ge \frac { 1 }{ \mu } \)

or \(1-{ cos }^{ 2 }r\le 1-\frac { 1 }{ { \mu }^{ 2 } } \Rightarrow { sin }^{ 2 }r\le 1-\frac { 1 }{ { \mu }^{ 2 } } \left[ \therefore 1-{ cos }^{ 2 }r={ sin }^{ 2 }r \right] \)

Let d be the diameter of the disc. The sopt shall be invisible, if the incident rays OA and OB suffer total internal reflection.
Let i be the angle of incidence

Using relationship between refractive index and critical angle, then
\(\sin { i=\frac { 1 }{ \mu } }\)
Using geometry and trigonometry,
Now,
\(\frac { d/2 }{ h } =\tan { i } \Rightarrow \frac { d }{ 2 } =h\tan { i } =h[\sqrt { { \mu }^{ 2 }-1{ ] }^{ -1 } }\)
\( [\because \ From \ the \ figure,\ tani=\frac { 1 }{ \sqrt { { \mu }^{ 2 } } -1 } ]\)
\( d=\frac { 2h }{ \sqrt { { \mu }^{ 2 } } -1 } \)
This is required expression of d.