The refractive index of vacuum, n₁ = 1
The wavelength in vacuum, λ₁ = 5893 Å.
The speed in vacuum, c = v₁ = 3 x 10⁸ m s^–1
The refractive index of water, n₂ = 1.33
The wavelength of light in water, λ₂
The speed of light in water, v₂
(a) The equation relating the wavelength and refractive index is,
\(\cfrac { { \lambda }_{ 1 } }{ \lambda _{ 2 } } =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \) 
Rewriting, \({ \lambda }_{ 2 }=\cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \times { \lambda }_{ 1 }\)
Substituting the values,
 \({ \lambda }_{ 2 }=\cfrac { 1 }{ 1.33 } \times 5893\overset { o }{ A } =4431\overset { o }{ A } \) 
\({ \lambda }_{ 2 }=4431\overset { o }{ A } \) 
(b) The equation relating the speed and refractive index is,
\(\cfrac { { v }_{ 1 } }{ { v }_{ 2 } } =\cfrac { { n }_{ 2 } }{ { n }_{ 1 } } \)
Rewriting, \({ v }_{ 2 }=\cfrac { { n }_{ 1 } }{ { n }_{ 2 } } \times { v }_{ 1 }\) 
Substituting the values,
\({ v }_{ 2 }=\cfrac { 1 }{ 1.33 } \times 3\times { 10 }^{ 8 }=2.256\times { 10 }^{ 8 }\) 
v₂ = 2.256 x 10⁸ ms^-1
(c) Frequency of light in vacuum is,
\({ v }_{ 1 }=\cfrac { c }{ { \lambda }_{ 1 } } \) 
Substituting the values,
\({ v }_{ 1 }=\cfrac { 3\times { 10 }^{ 8 } }{ 5893\times { 10 }^{ -10 } } =5.091\times { 10 }^{ 14 }Hz\)
Frequency of light in water is, \({ v }_{ 2 }=\cfrac { v }{ { \lambda }_{ 2 } } \)
Substituting the values,
 \(\\ { v }_{ 2 }=\cfrac { 2.256\times { 10 }^{ 8 }{ ms }^{ -1 } }{ 4431\times { 10 }^{ -10 } } =5.091\times { 10 }^{ 14 }Hz\)
The results show that the frequency remains same in all media.

Given, n_v = 1.632; n_R = 1.613; n_G = 1.620
Equation for dispersive power is,
\(\omega =\cfrac { \left( { n }_{ v }-{ n }_{ g } \right) }{ \left( { n }_{ G }-1 \right) } \) 
Substituting the values,
\(\omega =\cfrac { 1.632-1.613 }{ 1.620-1 } =\cfrac { 0.019 }{ 0.620 } =0.0306\) 
The dispersive power of the prism is,\(\\ \omega =0.0306\)

Given, A = 60°; D = 37°
Equation for refractive index is,
\(n=\cfrac { \sin\left( \frac { A+D }{ 2 } \right) }{ \sin\left( \frac { A }{ 2 } \right) } \) 
Substituting the values,
\(n=\cfrac { \sin\left( \frac { 60^{ o }+37^{ o } }{ 2 } \right) }{ \sin\left( \cfrac { { 60 }^{ o } }{ 2 } \right) } =\cfrac { \sin\left( 48.5^{ o } \right) }{ \sin\left( { 30 }^{ o } \right) } =1.5\)
The refractive index of the material of the prism is, n = 1.5 

Since, the prism is equilateral, A = 60^o;
Given, i₁ = 30^o;i₂ = 75^o
Equation for angle of deviation, d = i₁ + i₂ – A
Substituting the values, d = 30°+ 75°– 60°= 45°
The angle of deviation produced d = 45^o

The given situation is shown in the figure
Given, A = 60^o ; i₁ = 0^o; i₂ = 90^o
(a) Equation for angle of deviation,
d = i₁ + i₂ - A
Substituting the values
d = 0^o + 90^o- 60^o = 30^o
The angle of deviation produced is, d = 30°
(b) The light inside the prism must be falling on the second face at critical angle as it graces the boundary. i_c = 90° – 30° = 60°
Equation for critical angle is, \(\sin i_{c}=\frac{1}{n}\) 
\(n=\frac{1}{\sin i_{c}} ; \quad n=\frac{1}{\sin 60^{\circ}}=\frac{1}{\sqrt{3} / 2}=\frac{2}{\sqrt{3}}=1.15\) 
The refractive index of the material of the prism is, n = 1.15