Lesson: Vision and the Eye

In this lesson, we will learn how to calculate the power, near and far points, and accommodation of both normal eyes and those requiring correction.

Sample Question Videos

  • 01:31

Worksheet: Vision and the Eye • 20 Questions • 1 Video

Q1:

A nearsighted man cannot see objects clearly beyond 20 cm from his eyes. How close must he stand to a mirror in order to see what he is doing when he shaves?

Q2:

What is the power of the eye when viewing an object 50.0 cm away? The lens-to-retina distance is 2.00 cm.

Q3:

The power of a physician’s eyes is 53.0 D while examining a patient. How far from her eyes is the object that is being examined? The lens-to-retina distance is 2.00 cm.

Q4:

What is the near point of a person whose eyes have an accommodated power of 53.5 D? The lens-to-retina distance is 2.00 cm.

Q5:

A myopic person sees that her prescription for her glasses is 4 . 0 0 D. What is her far point if the glasses are 1.75 cm from her eyes? The lens-to-retina distance is 2.00 cm.

Q6:

The power for normal distant vision is 50.0 D. A severely myopic patient has a far point of 5.00 cm. By how many diopters should the power of his eye be reduced in laser vision correction to obtain normal distant vision for him? The lens-to-retina distance is 2.00 cm.

Q7:

A very myopic man has a far point of 20.0 cm. What power eyeglasses held 1.50 cm from his eyes will correct his distant vision? The lens-to-retina distance is 2.00 cm.

Q8:

The print in many books averages 3.50 mm in height. How high is the image of the print on the retina when the book is held 30.0 cm from the eye? The lens-to-retina distance is 2.00 cm.

Q9:

A mother sees that her child’s glasses prescription is 0.750 D. What is the child’s near point if the glasses are held 2.20 cm from the child’s eyes?

Q10:

The power for normal close vision is 54.0 D. In a vision-correction procedure, the power of a patient’s eye is increased by 3.00 D. Assuming that this produces normal close vision, what was the patient’s near point before the procedure? The lens-to-retina distance is 2.00 cm.

Q11:

The far point of a myopic administrator is 50.0 cm. The lens-to-retina distance is 2.00 cm.

What is the relaxed power of his eyes?

If he has the normal 8 . 0 0 % ability to accommodate, what is the closest object he can see clearly?

Q12:

People who do very detailed work close up, such as jewelers, often can see objects clearly at much closer distance than the normal 25 cm. The lens-to-retina distance is 2.00 cm.

What is the power of the eyes of a woman who can see an object clearly at a distance of only 8.00 cm?

What is the image size of an object 1.00 mm long, such as lettering inside a ring, held at a distance of 8.00 cm?

What would the size of the image be if an object 1.00 mm long was held at the normal 25.0 cm distance?

Q13:

An object with a height of 1.50 cm is held at a distance of 3.00 cm from a person’s cornea. A reflected image of the object on the cornea has a height of 0.157 cm.

What is the magnification of the reflected image?

How far behind the cornea is the image formed?

Find the radius of curvature of the convex mirror formed by the cornea.

Q14:

The cornea and eye lens have focal lengths of 2.3 and 6.4 cm respectively. Find the optical power of the eye.

Q15:

A very myopic man has a far point of 23.0 cm. What power contact lens, placed directly onto the eye, will correct his distant vision?

Q16:

Calculate the power of the lens in a human eye when focusing on an object 4.0 m away from it.

Q17:

A myopic person sees that her contact lens prescription is −5.0 D. What is her far point?

Q18:

A student’s eyes, while reading from a board, have a power of 53.0 D. How far is the board from the student’s eyes, assuming that the student’s eyes have an image distance of 2.00 cm?

Q19:

What is the far point of a person whose eyes have a relaxed power of 52 D?

Q20:

What range of magnification is possible with an 8.3 cm focal length converging lens?

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