Answer:
The range of powers is [tex]- 5 \ D \le P \le - 2.667\ D[/tex]
Explanation:
From the question we are told that
The far point of the left eye is [tex]n_f = 20 cm[/tex]
The near point of the left eye is [tex]n = 15cm[/tex]
The near point with the glasses on is [tex]n_g =25 \ cm[/tex]
From these parameter we can see that with the glass on that for near point the
Object distance would be [tex]u = -25 \ cm[/tex]
Image distance would be [tex]v = -15 \ cm[/tex]
To obtain the focal length we would apply the lens formula which is mathematically represented as
[tex]\frac{1}{f} = \frac{1}{v} - \frac{1}{u}[/tex]
substituting values
[tex]\frac{1}{f} = \frac{1}{-15} - \frac{1}{-25}[/tex]
[tex]f = - \frac{75}{2} cm[/tex]
converting to meters
[tex]f = - \frac{75}{2} * \frac{1}{100}[/tex]
[tex]f = - \frac{75}{200} \ m[/tex]
Generally the power of the lens is mathematically represented as
[tex]P = \frac{1}{f}[/tex]
Substituting values
[tex]P = - \frac{200}{75} m[/tex]
[tex]P = - 2.667 \ D[/tex]
From these parameter we can see that with the glass on that for far point the
Object distance would be [tex]u_f = - \infty \ cm[/tex]
Image distance would be [tex]v_f = -20 \ cm[/tex]
To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as
[tex]\frac{1}{f_f} = \frac{1}{v_f} - \frac{1}{u_f}[/tex]
substituting values
[tex]\frac{1}{f} = \frac{1}{-20} - \frac{1}{- \infty}[/tex]
[tex]\frac{1}{f} = \frac{1}{-20} - 0[/tex]
[tex]f_f = \frac{20}{1} \ cm[/tex]
converting to meters
[tex]f_f = - \frac{20}{1} * \frac{1}{100}[/tex]
Generally the power of the lens is mathematically represented as
[tex]P = \frac{1}{f_f}[/tex]
Substituting values
[tex]P = - \frac{100}{20} m[/tex]
[tex]P = - 5 \ D[/tex]
This implies that the range of powers of the lens in his glass is
[tex]- 5 \ D \le P \le - 2.667\ D[/tex]
