The image can be located by graphical construction. To locate the point of an image, a minimum of two rays must meet at that point.
(i) A ray parallel to the principal axis after reflection will pass through or appear to pass through the principle focus. (Figure(a))

(ii) A ray passing through or appear to pass through the principal focus after reflection will travel parallel to the principal axis. (Figure(b) )

(iii) A ray passing through the center of curvature retraces its path after reflection as it is a case of normal incidence. (Figure(c))

(iv) A ray falling on the pole will get reflected as per the law of reflection keeping the principal axis as normal. (Figure( d))

\(\cfrac { 1 }{ f } =\left( { n }_{ 2 }-1 \right) \left( \cfrac { 1 }{ { R }_{ 1 } } -\cfrac { 1 }{ { R }_{ 2 } } \right) \) 
= \(\left( 1.5-1 \right) \left( \cfrac { 1 }{ 20 } -\cfrac { 1 }{ -20 } \right) =\cfrac { 1 }{ 20 } \) 
or f = 20 cm
Using \(\cfrac { 1 }{ v } -\cfrac { 1 }{ -f } =\cfrac { 1 }{ f } \) ,we get
\(\cfrac { 1 }{ v } -\cfrac { 1 }{ -15 } =\cfrac { 1 }{ 20 } \)
v = -60 cm.

From the lens maker's formula,
\(\cfrac { 1 }{ f } =\left( \mu -1 \right) \left( \frac { 1 }{ { R }_{ 1 } } -\frac { 1 }{ { R }_{ 2 } } \right) \) 
Here, μ = 1.5; R₁ = R₂ and  =-R
\(\therefore \cfrac { 1 }{ f } =\left( 1.5-1 \right) \left( \cfrac { 1 }{ R } -\cfrac { 1 }{ -R } \right) =0.5\times \cfrac { 2 }{ R } =\cfrac { 1 }{ R } \) 
or f = R.

\(sin{ i }_{ c }=\cfrac { 1 }{ \mu } =\cfrac { 1 }{ 1.5 } =0.667\\ \) 
or i_c= 41.8^o
Deviation = 90^o - i = 90^o - 41.8^o = 48.2^o
This si the maximum attainable deviation in refraction. So, the given data favours total internal reflection.
In reflection, deviation = 180^o - 2i
or 90^o = 180^o - 2i or 2i = 90^o or i = 45^o
or i = 45^o

For minimum deviation position
\({ \mu }_{ red }=\cfrac { sin\left( \frac { A+{ \delta }_{ red } }{ 2 } \right) }{ sin\frac { A }{ 2 } } \) 
or \(sin\left( \cfrac { A+{ { \delta }_{ red } } }{ 2 } \right) ={ n }_{ red }\) 
\(sin\cfrac { 2 }{ A } =1.622\times 0.5=0.811\) 
\(\therefore \cfrac { 60+{ \lambda }_{ red } }{ 2 } ={ 54 }^{ o }12'\) 
\({ \delta }_{ red }={ 108 }^{ o }24'-{ 60 }^{ o }={ 48 }^{ o }24'\)
Similarly, \(sin\left( \cfrac { A+{ \delta }_{ viloet } }{ 2 } \right) =1.663\times 0.5'\) 
= 0.8315 or \(\cfrac { { 60 }^{ o }+{ \delta }_{ viloet } }{ 2 } =56^{ o }15'\) 
\({ \delta }_{ viloet }=112^{ o }30'-{ 60 }^{ ' }={ 52 }^{ o }30'\) 
\(\therefore { \delta }_{ violet }-{ \delta }_{ red }=\left( 52-30' \right) -\left( 48^{ o }24' \right) ={ 4 }^{ o }6'\) 
It is not advisable to use the formula,
\({ \delta }_{ v }-{ \delta }_{ r }=\left( { \mu }_{ v }-{ \mu }_{ r } \right) \) in the above solution.