Answer:
The value is [tex]p = - 2.63 \ Diopters[/tex]
Explanation:
From the question we are told that
The value of the far point is [tex]a = 40 \ cm = 0.4 \ m[/tex]
The distance of the lens to the eye is [tex]b = 2 \ cm = 0.02[/tex]
Generally
[tex]1 Diopter = > 1 m^{-1}[/tex]
Generally the power spectacle lens needed is mathematically represented as
[tex]p = \frac{1}{d_o } + \frac{1}{d_i}[/tex]
Here [tex]d_o[/tex] is the object distance which for a near sighted person is [tex]d_o = \infty[/tex]
And [tex]d_i[/tex] is the image distance which is evaluated as
[tex]d_i = b - a[/tex]
=> [tex]d_i = 0.02 - 0.4[/tex]
=> [tex]d_i = -0.38 \ m[/tex]
So
[tex]p = \frac{1}{\infty } + \frac{1}{-0.38}[/tex]
=> [tex]p = 0 - 2.63[/tex]
=> [tex]p = - 2.63 \ Diopters[/tex]
