Answer:
The near point is [tex]n =44.8 \ cm[/tex]
Explanation:
From the question we are told that
The power is [tex]P = 1.50[/tex]
The distance from the eye is [tex]k = 1.8 \ cm[/tex]
The distance of the book from the eye is [tex]z = -28 \ cm[/tex]
Generally the focal length of the glasses is
[tex]f = \frac{1}{P}[/tex]
=> [tex]f = \frac{1}{1.50 }[/tex]
=> [tex]f = 0.667 \ m[/tex]
=> [tex]f = 66.7 \ cm[/tex]
The object distance is evaluated as
[tex]u = z + k[/tex]
=> [tex]u = -28 + 1.8[/tex]
=> [tex]u = -26.2 \ cm[/tex]
The image distance is evaluated from lens formula as
[tex]\frac{1}{v} = \frac{1}{f} + \frac{1}{u}[/tex]
=> [tex]\frac{1}{v} = \frac{1}{66.7} + \frac{1}{-26.2}[/tex]
=> [tex]v=- \frac{1}{0.0232}[/tex]
=> [tex]v=- 43 \ cm[/tex]
The near point is evaluated as
[tex]n = -v + k[/tex]
=> [tex]n =-(-43) + 1.8[/tex]
=> [tex]n =44.8 \ cm[/tex]
