Given, n₁ = 1.68, n₂ = 1.44, n₃ = 1
Acceptance angle, \(\\ { i }_{ a }={ sin }^{ -1 }\left( \sqrt { { n }_{ 1 }^{ 2 }-{ n }_{ 2 }^{ 2 } } \right) \) 
\({ i }_{ a }={ sin }^{ -1 }\left( \sqrt { \left( 1.68 \right)^2 -\left( 1.44 \right) ^{ 2 } } \right) ={ sin }^{ -1 }\left( 0.865 \right) \) 
\({ i }_{ a }\simeq { 60 }^{ o }\) 
If there is no cladding then, n₂ = 1
Acceptance angle, \({ i }_{ a }={ sin }^{ -1 }\left( \sqrt { { n }_{ 1 }^{ 2 }-1 } \right) \)
\({ i }_{ a }={ sin }^{ -1 }\left( \sqrt { \left( 1.68 \right) ^{ 2 }-1 } \right) ={ sin }^{ -1 }\left( 1.35 \right) \) 
sin⁻¹(more than 1) is not possible. But, this includes the range 0^o to 90^o. Hence, all the rays entering the core from flat surface will undergo total internal reflection.
Note: If there is no cladding then there is a condition on the refractive index (n₁) of the core
\({ i }_{ a }={ sin }^{ -1 }\left( \sqrt { { n }_{ 1 }^{ 2 }-1 } \right) \) 
Here, as per mathematical rule,
\(\left( { n }_{ 1 }^{ 2 }-1 \right) \le 1\) or \(\left( { n }_{ 1 }^{ 2 } \right) \le 2\) or \({ n }_{ 1 }\le \sqrt { 2 } \)
Hence, in air (no cladding) the refractive index n₁ of the core should be,\({ n }_{ 1 }\le 1.414\)

When seen from (air medium) on top, the coin will still appear to be at the bottom with each medium appearing to have shrunk with respect to the air medium outside. This situation is illustrated below.

The equations for apparent depth for each medium is,
\({ d' }_{ 1 }=\cfrac { { d }_{ 1 } }{ { n }_{ 1 } } ;{ d }_{ 2 }^{ ' }=\cfrac { { d }_{ 2 } }{ { n }_{ 2 } } ;{ d }_{ 3 }^{ ' }=\cfrac { { d }_{ 3 } }{ { n }_{ 3 } } \) 
\({ d }^{ ' }={ d }_{ 1 }^{ ' }+{ d }_{ 2 }^{ ' }+{ d }_{ 3 }^{ ' }=\cfrac { { d }_{ 1 } }{ n_{ 1 } } +\cfrac { { d }_{ 2 } }{ { n }_{ 2 } } +\cfrac { { d }_{ 3 } }{ n_{ 3 } } \) 
\(d'=\cfrac { 30 }{ 1.3 } +\cfrac { 1.6 }{ 1.4 } +\cfrac { 30 }{ 1.5 } =23.1+11.4+13.3\) 
d' = 47.8 cm

Given, u = –40 cm, R = –20 cm, n₁ = 1 and n₂ = 1.33
Equation for single spherical surface is
\(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
Substituting the values
\(\cfrac { 1.33 }{ v } -\cfrac { 1 }{ -40 } =\cfrac { \left( 1.33-1 \right) }{ -20 } ;\cfrac { 1.33 }{ v } +\cfrac { 1 }{ 40 } =\cfrac { \left( 0.33 \right) }{ -20 } \) 
\(\cfrac { 1.33 }{ v } =\cfrac { \left( 0.33 \right) }{ 20 } -\cfrac { 1 }{ 40 } ;\) 
\(\cfrac { 1.33 }{ v } =\cfrac { 0.66-1 }{ 40 } =\cfrac { 1.66 }{ 40 } \) 
\(v=-40\times \cfrac { 1.33 }{ 1.66 } =-32.0cm\)
The equation for magnification is,\(m=\cfrac { { h }_{ 2 } }{ { h }_{ 1 } } =\cfrac { { { n }_{ 1 }v } }{ { n }_{ 2 }u } \) 
\(\cfrac { { h }_{ 2 } }{ 1.0 } =\cfrac { \left( 1.0 \right) \times \left( -32 \right) }{ \left( 1.33 \right) \times \left( -40 \right) } =0.6cm\)  (or) h₂ = 0.6cm
The erect virtual image of height 0.6 cm is formed at 32.0 cm to the left of the single spherical surface.

Case i) \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
applying sign convention, u = –10 cm, R = 15 cm
\(\cfrac { 2 }{ v } -\cfrac { 1 }{ -10 } =\cfrac { \left( 2-1 \right) }{ 15 } ;\cfrac { 2 }{ v } +\cfrac { 1 }{ 15 } =\cfrac { \left( 1 \right) }{ 10 } \) 
∴ =− 60 cm
[a virtual image is formed 60 cm, to the left of the surface]

Case ii) \(\cfrac { { n }_{ 2 } }{ v } -\cfrac { { n }_{ 1 } }{ u } =\cfrac { \left( { n }_{ 2 }-{ n }_{ 1 } \right) }{ R } \)
applying sign convention, u = –30 cm, R = 15 cm
\(\cfrac { 2 }{ v } -\cfrac { 1 }{ -30 } =\cfrac { \left( 2-1 \right) }{ 15 } ;\cfrac { 2 }{ v } +\cfrac { 1 }{ 30 } =\cfrac { \left( 1 \right) }{ 15 } \)
∴ = 60 cm [a real image is formed 60 cm, to the right of the surface]

Given, thickness of the slab, t = 0.25 m, refractive index, n = 1.5, angle of incidence, i = 60^o.
Using Snell’s law, 1 sin i = n sin r
\(sinr=\cfrac { sini }{ n } =\cfrac { sin60^o }{ 1.5 } =0.58\) 
\(r={ sin }^{ -1 }(0.58)=35.25^{ 0 }=35^o15'0''\) 
Lateral displacement is, \(L=t\left( \cfrac { sin\left( i-r \right) }{ cos\left( r \right) } \right) \) 
\(L=\left( 0.25 \right) \times \left( \cfrac { sin\left( 60-35.25 \right) }{ cos\left( 35.25 \right) } \right) =0.1281m\)
The  lateral displacement is, L = 12.81 cm

Given, n = 4/3, d = 10 m.
Radius of illumination, \(R=\cfrac { d }{ \sqrt { { n }^{ 2 }-1 } } \) 
\(R=\cfrac { 10 }{ \sqrt { \left( 4/3 \right) ^{ 2 }- } 1 } =\cfrac { 10\times 3 }{ \sqrt { 16-9 } } \) 
\(R=\cfrac { 30 }{ \sqrt { 7 } } =11.32cm\) 
To find the critical angle,
\({ i }_{ c }={ sin }^{ -1 }\left( \cfrac { 1 }{ n } \right) \) 
\({ i }_{ c }={ sin }^{ -1 }\left( \cfrac { 1 }{ 4/3 } \right) ={ sin }^{ -1 }\left( \cfrac { 3 }{ 4 } \right) =48.6^{ o }\)
The total angle of view of the cone is, \({ 2i }_{ c }=2\times { 48.6 }^{ 0 }={ 97.2 }^{ 0 }\)