Question

A person's left eye has a near point of 43 cm and requires
eyeglasses to correct their vision.

a) What would the prescription for the left eye?

b) A 9.0 mm base curve (radius of curvature) will best fit the
person's cornea. Using the material with index of refractive 1.422,
what should be the radius of curvature (include the correct sign)
fo the other side of the lens?

c) What is their new far point after wearing the contact lens?

Answer #1

A) for an healthy eye near point must be at 25 cm so to choose lens we have to shift this near point from 43 cm to 25cm

So, object distance p=- 43cm

and image distance q= -25 cm

So from lens maker formula

So focal length

f = -59.72 cm

So power of lens = 1/f = -0.0167 Dioptre

2) So focal length can be written as

n = 1.422

R1 = 9 cm

R2= -R

So solving for R2 we get

R2 = -13.97 cm for the other side

C) so near far point is now at 25 cm from left eye

A person's prescription for his new bifocal eyeglasses calls for
a refractive power of -0.0600 diopters in the distance-vision part
and a power of 1.05 diopters in the close-vision part. Assuming the
glasses rest 2.00 cm from his eyes and that the corrected
near-point distance is 25.0 cm, determine the near and far points
of this person's uncorrected vision.

A. Where is the near point of an eye for which a contact lens
with a power of +2.85 diopters is prescribed? B. Where is the far
point of an eye for which a contact lens with a power of –1.10
diopters is prescribed for distant vision?

(a) Where is the near point of an eye for which a contact lens
with a power of +2.95 diopters is prescribed?
cm (in front of the eye)
(b) Where is the far point of an eye for which a contact lens with
a power of -1.35 diopters is prescribed for distant vision?
cm (in front of the eye)

A person is farsighted and has near point of 50 cm. Retina is at
a distance L = 1.7 cm. When wearing contact lenses and reading a
book 30 cm in front of the eye, what is the object distance for the
lenses? What is the desired image distance? Is the magnitude of the
image distance a minimum, maximum, or the only acceptable image
distance? Find the focal length and refractive power of the contact
lens that will fix this...

The physical properties of the relaxed eye are: distance from
front surface of cornea to front surface of the lens: 3.6
mm, thickness of lens: 3.6 mm, cornea (r1): 8
mm, anterior surface of lens (r2):10.0 mm,
posterior surface of lens (r3): 6.0 mm. The refractive
index of aqueous and vitreous humor and lens are 1.34 and 1.41,
respectively. With these properties, the distance from the cornea
to the retina is 25 mm and an object infinitely far away
is...

The cornea behaves as a thin lens of focal length approximately
1.80 cm , although this varies a bit. The material of which it is
made has an index of refraction of 1.38, and its front surface is
convex, with a radius of curvature of 5.00 mm .
(Note: The results obtained here are not strictly
accurate, because, on one side, the cornea has a fluid with a
refractive index different from that of air.)
Part A: If this focal...

People with normal vision cannot focus their eyes underwater if
they aren't wearing a face mask or goggles and there is water in
contact with their eyes. In a simplified model of the human eye,
the aqueous and vitreous humors and the lens all have a refractive
index of 1.40, and all the refraction occurs at the cornea, whose
vertex is 2.60 cm from the retina. Part A With the simplified model
of the eye, what corrective lens (specified by...

A. The human eye 1. Whenever a normal eye forms an image, the
image distance will always equal the distance from the cornea and
eye lens to the retina (~25 mm), regardless of how far away the
object is located. Explain why the image distance cannot change. 2.
If the image distance must change, then what intrinsic property of
the eye lens must change in order for the eye to focus on objects
at different distances? Hint: read the Introduction....

The cornea of the eye has a radius of curvature of approximately
0.40 cm , and the aqueous humor behind it has an index of
refraction of 1.35. The thickness of the cornea itself is small
enough that we shall neglect it. The depth of a typical human eye
is around 25.0 mm .
a. What would have to be the radius of curvature of the cornea
so that it alone would focus the image of a distant mountain on...

The near point of a patient's eye is 54.2 cm.
(a) What should be the refractive power ℛ of a
corrective lens prescribed to enable the patient to clearly see an
object at 23.0 cm?
ℛ = dpt
(b) When using the new corrective glasses, the patient can see an
object clearly at 26.8 cm but not at 23.0 cm. By how many diopters
did the lens grinder miss the prescription?
? = dpt

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 22 minutes ago

asked 37 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago