The minimum distance the person could see clearly is, y = 75 cm.
The lens should have a focal length of,
\(f=\cfrac { y\times 25cm }{ y-25cm } \) 
\(f=\cfrac { 75cm\times 25cm }{ 75cm-25cm } =37.5cm\) 
It is a convex (or) converging lens.
The power of the lens is,
\(P=\cfrac { 1 }{ 0.375m } =2.67\ D\)

Brewster’s law, tan i_p = n
For glass, tani_p = 1.5 ; i_p = tan^-11.5 ; i_p= 56.3^o
For water, tani_p= 1.33; i_p= tan^-1 = tan^-1 1.33; i_p = 53.1^o

As the intensity of the unpolarised light falling on the first polaroid is I, the intensity of polarized light emerging from will be, \({ I }_{ 0 }=\left( \cfrac { 1 }{ 2 } \right) \) 
Let I' be the intensity of light emerging from the second polaroid.
Malus’ law, I' = I_o = cos²θ
Substituting,
\({ I }^{ ' }=\left( \cfrac { 1 }{ 2 } \right) { cos }^{ 2 }\left( { 30 }^{ o } \right) =\left( \cfrac { 1 }{ 2 } \right) \left( \cfrac { \sqrt { 3 } }{ 2 } \right) ^{ 2 }=1\cfrac { 3 }{ 8 } \) 
\(I'=\left( \cfrac { 3 }{ 8 } \right) I\)

f_o = 125 cm; f_e = 2 cm; m = ? L = ?; θ_i = ?
(a) Equation for magnification of telescope,
\(m=\cfrac { { f }_{ o } }{ { f }_{ e } } \) 
Substituting, \(m=\cfrac { 125 }{ 2 } =62.5\) 
(b) Equation for approximate length of telescope, L = f_o+ f_e
Substituting, L = 125 + 2 = 127 cm = 1.27 m
(c) Equation for angular magnification,\(m=\cfrac { { \theta }_{ 1 } }{ { \theta }_{ 0 } } \) 
Rewriting, \({ \theta }_{ 1 }=m\times { \theta }_{ 0 }\) 
Substituting,
\({ \theta }_{ i }=62.5\times 1'=62.5'=\cfrac { 62.5 }{ 60 } =1.04^{ o }\) = 1^o2'30''

D = 25 cm;
The magnifying lens must be a convex lens of positive focal length
f = 5 cm;
For closest object distance u', the image distance, v is, –25 cm. (near point, v = –D)
For farthest object distance u', the corresponding image distance, v' is infinity.
(a) To find closest distance between lens and book, we can use lens equation,\(\cfrac { 1 }{ v } -\cfrac { 1 }{ u } =\cfrac { 1 }{ f } \) 
Rewriting for closest object distance \(\cfrac { 1 }{ u } =\cfrac { 1 }{ v } -\cfrac { 1 }{ f } \) 
Substituting,
\(\cfrac { 1 }{ u } =\cfrac { 1 }{ -25 } -\cfrac { 1 }{ 5 } =\cfrac { 1 }{ 25 } =\cfrac { 1 }{ 5 } \left( \cfrac { -1-5 }{ 25 } \right) =-\cfrac { 6 }{ 25 } \)
\(u=-\cfrac { 25 }{ 6 } =4.167cm\)
The closest distance between the lens and the book is, u = –4.167 cm
To find farthest object distance, lens equation is, \(\cfrac { 1 }{ v' } -\cfrac { 1 }{ u' } =\cfrac { 1 }{ f' } \)
Rewriting for farthest object distance,\(\cfrac { 1 }{ u' } =\cfrac { 1 }{ v' } -\cfrac { 1 }{ f' } \)
Substituting,\(\cfrac { 1 }{ u' } =\cfrac { 1 }{ \infty } -\cfrac { 1 }{ 5 } ;u=-5cm\) 
The farthest distance at which the person can keep the book is, u' = -5 cm.
(b) To find magnification in near point focusing, \(m=1+\cfrac { D }{ f } =1+\cfrac { 25 }{ 5 } =6\)
To find magnification in normal focusing, \(m=\cfrac { D }{ f } =\cfrac { 25 }{ 5 } =5\)