Question

A patient can't see objects closer than 32.0 cm and wishes to clearly see objects that are 20.0 cm from his eye.

(a) Is the patient nearsighted or farsighted?

nearsighted farsighted

(b) If the eye–lens distance is 1.96 cm, what is the minimum object
distance *p* from the lens? (Give your answer to at least
three significant digits.)

cm

(c) What image position with respect to the lens will allow the
patient to see the object? (Give your answer to at least three
significant digits.)

cm

(d) Is the image real or virtual? Is the image distance *q*
positive or negative? (Select all that apply.)

real image virtual image distance is positive distance is negative

(e) Calculate the required focal length.

cm

(f) Find the power of the lens in diopters.

diopters

(g) If a contact lens is to be prescribed instead, find *p*,
*q*, and *f*, and the power of the lens.

p |
= | cm | (Give your answer to at least three significant digits.) |

q |
= | cm | (Give your answer to at least three significant digits.) |

f |
= | cm | |

P |
= | diopters |

Answer #1

(a). The person is suffering from farsightedness.

(b). From the lens, the minimum distance of the object should be -(20-1.96)=-18.4cm.

(c). The desired image position from the lens is -(32-1.96)=-30.4cm.

(d). The image is virtual. Sign of p,q f depends on the convention. In my convention, I put objects to the left of the lens, and write all distances to the left as negative and all distances to the right as positive. So q will come out to be negative.

(e).

p=-18.4cm,

q=-30.4cm.

(f).

(g). p=-18.4cm,

q=-30m,

f=46.613cm,

P=0.021D.

Contact lenses are placed right on the eyeball, so the distance
from the eye to an object (or image) is the same as the distance
from the lens to that object (or image). A certain person can see
distant objects well, but his near point is 50.0 cm from his eyes
instead of the usual 25.0 cm .
Part A
Is this person nearsighted or farsighted?
Is this person nearsighted or farsighted?
This person is nearsighted.
This person is farsighted....

A converging lens has a focal length of 10.0 cm . For each of
two objects located to the left of the lens, one at a distance of
s1 = 20.5 cm and the other at a distance of s2 = 5.00 cm ,
determine the following.
A)Determine the image position.Express your answer in
centimeters separated by a comma to three significant figures.
B)Determine the magnification. Express your answer in
centimeters separated by a comma to three significant figures.
C)Determine...

A patient's far point is 135 cm and her near point is 15.0
cm.
In what follows, we assume that we can model the eye as a simple
camera, with a single thin lens forming a real image upon the
retina. We also assume that the patient's eyes are identical, with
each retina lying 1.95 cm from the eye's "thin lens."
(a)
What is the power, P, of the eye when focused upon the
far point? (Enter your answer in...

The uncorrected eye.
For the following questions, assume that the distance between the
eye lens and the retina is 1.70 cm. In other words, since the image
is always formed on the retina, the distance between the lens and
the image is always 1.70 cm. Also note that, as is seen in the ray
diagram, since the eye lens is converging and the image is on the
opposite side of the lens compared to the object, the image is
always...

A person's right eye can see objects clearly only if they are
between 26 cm and 88 cm away.
Part A What power of contact lens is required so that objects
far away are sharp? Express your answer using two significant
figures.
Part B What will be the near point with the lens in place?
Express your answer using two significant figures and include the
appropriate units.

A converging lens has a focal length of 10.0 cm . For each of
two objects located to the left of the lens, one at a distance of
s1 = 20.5 cm and the other at a distance of s2 = 5.00 cm ,
determine the following.
Determine the image position.Express your answer in centimeters
separated by a comma to three significant figures.
is the image real or virtual

Contact lenses are placed on the eyeball, so the distance from
the eye to an object (or an image) is the same as the distance from
the lens to the object (or the image). If a person can see distant
objects well, but has a nearpoint of 53.0 cm instead of the usual
25.0 cm, what power of contact lenses are needed (in diopters)?
The same person now wants to wear glasses instead; they are 1.90
cm in front of...

People who do very detailed work close up, such as jewellers,
often can see objects clearly at much closer distance than the
normal 25 cm.
Part (a) What is the power in D of the eyes of a woman who can
see an object clearly at a distance of only 7.5 cm? Assume the
distance to her retina from the lens in her eye is 2.00
cm.
Part (b) What is the size in mm of an image of...

A person is both myopic and hyperopic. He can see objects
clearly only if they are between 31.3 cm c m and 80.0 cm c m . Part
A What power of contact lens is required to see a distant object
clearly? Express your answer using three significant figures. Part
B What power of contact lens is required to read at the normal near
point of 25 cm?

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....

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 26 seconds ago

asked 6 minutes ago

asked 8 minutes ago

asked 35 minutes ago

asked 36 minutes ago

asked 39 minutes ago

asked 52 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago