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I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limited range of accommodation. My glasses have progressive lenses. The lenses are diverging lenses, with a negative power, but the magnitude of the power varies from a maximum at the top, for distance vision, to a minimum at the bottom, for close vision. With no correction, for my left eye, my near point is 15 cm, my far point is 20 cm. With my glasses on, my near point is about 25 cm, and my far point is infinity. What is the range of powers of the lens in my glasses?

User Axel Meier
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1 Answer

5 votes

Answer:

The range of powers is
- 5 \ D \le P \le - 2.667\ D

Step-by-step explanation:

From the question we are told that

The far point of the left eye is
n_f = 20 cm

The near point of the left eye is
n = 15cm

The near point with the glasses on is
n_g =25 \ cm

From these parameter we can see that with the glass on that for near point the

Object distance would be
u = -25 \ cm

Image distance would be
v = -15 \ cm

To obtain the focal length we would apply the lens formula which is mathematically represented as


(1)/(f) = (1)/(v) - (1)/(u)

substituting values


(1)/(f) = (1)/(-15) - (1)/(-25)


f = - (75)/(2) cm

converting to meters


f = - (75)/(2) * (1)/(100)


f = - (75)/(200) \ m

Generally the power of the lens is mathematically represented as


P = (1)/(f)

Substituting values


P = - (200)/(75) m


P = - 2.667 \ D

From these parameter we can see that with the glass on that for far point the

Object distance would be
u_f = - \infty \ cm

Image distance would be
v_f = -20 \ cm

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as


(1)/(f_f) = (1)/(v_f) - (1)/(u_f)

substituting values


(1)/(f) = (1)/(-20) - (1)/(- \infty)


(1)/(f) = (1)/(-20) - 0


f_f = (20)/(1) \ cm

converting to meters


f_f = - (20)/(1) * (1)/(100)

Generally the power of the lens is mathematically represented as


P = (1)/(f_f)

Substituting values


P = - (100)/(20) m


P = - 5 \ D

This implies that the range of powers of the lens in his glass is


- 5 \ D \le P \le - 2.667\ D

User TariqN
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