The ray diagram showing the image formation is given below:
 
O is at 2f of lens, so it will form image at 2f on other side, i.e. 60 cm from lens. Hence, position of object from mirror is at (60 - 15) cm = 45 cm behind the mirror.
For mirror,
\(f=+10cm,\) 
\( u=+45cm,\ v=?\)
As, \(\frac { 1 }{ v } +\frac { 1 }{ u } =\frac { 1 }{ f } \) 
\(\Rightarrow \ \frac { 1 }{ v } +\frac { 1 }{ 45 } =\frac { 1 }{ 10 } \) 
\( \therefore \ v=+\frac { 90 }{ 7 } cm\)   (behind be mirror)

By geometry, angle of incidence (i) at face AC for all three rays is  45\(^{0}\). Light suffers total internal reflection for which this angle of incidence is greater than critical angle.
 i>i_c, \(\Rightarrow\)sin i > sin i_c
or  sin 45\(^{0}\) > sin i_c \(\Rightarrow\) \(\frac { 1 }{ sin{ 45 }^{ 0 } } <\frac { 1 }{ sin{ i }_{ c } } \Longrightarrow \sqrt { 2 } <\mu \)
Total internal reflection takes place on AC for rays with \(\mu >\sqrt { 2 } =1.414,\) i.e., green and blue colour sukker. Total internal reflection whereas red undergoes refraction.

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.

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

(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 \({ P }_{ 3 }\)
 \(=\left( \frac { { I }_{ \circ } }{ 2 } cos^{ 2 }\beta \right) \left[ cos^{ 2 }\left( { 90 }^{ \circ }-\beta \right) \right] \)

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