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Chapter 26: Vision and Optical Instruments

21 problems · 1 figure

26-1 §26.1 source ↗

Unless otherwise stated, the lens-to-retina distance is 2.00 cm.

What is the power of the eye when viewing an object 50.0 cm away?

Answer

52.0 D

26-3 §26.1 source ↗

Unless otherwise stated, the lens-to-retina distance is 2.00 cm.

(a) 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?

(b) Compare the size of the print to the sizes of rods and cones in the fovea and discuss the possible details observable in the letters. (The eye-brain system can perform better because of interconnections and higher order image processing.)

Answer

(a) − 0 . 233 mm

(b) The size of the rods and the cones is smaller than the image height, so we can distinguish letters on a page.

26-5 §26.1 source ↗

Unless otherwise stated, the lens-to-retina distance is 2.00 cm.

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.

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

(b) What is the size of an image of a 1.00 mm object, such as lettering inside a ring, held at this distance?

(c) What would the size of the image be if the object were held at the normal 25.0 cm distance?

Answer

(a) +62.5 D

(b) –0.250 mm

(c) –0.0800 mm

26-6 §26.2 source ↗

What is the far point of a person whose eyes have a relaxed power of 50.5 D?

Answer

2.00 m

26-8 §26.2 source ↗

(a) A laser vision correction reshaping the cornea of a myopic patient reduces the power of his eye by 9.00 D, with a ±5.0% uncertainty in the final correction. What is the range of diopters for spectacle lenses that this person might need after LASIK procedure? (b) Was the person nearsighted or farsighted before the procedure? How do you know?

Answer

(a) ±0.45 D

(b) The person was nearsighted because the patient was myopic and the power was reduced.

26-10 §26.2 source ↗

What was the previous far point of a patient who had laser vision correction that reduced the power of her eye by 7.00 D, producing normal distant vision for her?

Answer

0.143 m

26-12 §26.2 source ↗

A student’s eyes, while reading the blackboard, have a power of 51.0 D. How far is the board from his eyes?

Answer

1.00 m

26-14 §26.2 source ↗

A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Answer

20.0 cm

26-16 §26.2 source ↗

A very myopic man has a far point of 20.0 cm. What power contact lens (when on the eye) will correct his distant vision?

Answer

–5.00 D

26-18 §26.2 source ↗

A myopic person sees that their contact lens prescription is –4.00 D. What is their far point?

Answer

25.0 cm

26-20 §26.2 source ↗

The contact lens prescription for a mildly farsighted person is 0.750 D, and the person has a near point of 29.0 cm. What is the power of the tear layer between the cornea and the lens if the correction is ideal, taking the tear layer into account?

Answer

–0.198 D

26-22 §26.2 source ↗

A teacher sees that a child’s contact lens prescription is 0.750 D. What is the child’s near point?

Answer

30.8 cm

26-24 §26.2 source ↗

The contact lens prescription for a nearsighted person is –4.00 D and the person has a far point of 22.5 cm. What is the power of the tear layer between the cornea and the lens if the correction is ideal, taking the tear layer into account?

Answer

–0.444 D

26-26 §26.4 source ↗

A microscope with an overall magnification of 800 has an objective that magnifies by 200. (a) What is the magnification of the eyepiece? (b) If there are two other objectives that can be used, having magnifications of 100 and 400, what other total magnifications are possible?

Answer

(a) 4.00

(b) 1600

26-28 §26.4 source ↗

(a) Where does an object need to be placed relative to a microscope for its 0.500 cm focal length objective to produce a magnification of –400? (b) Where should the 5.00 cm focal length eyepiece be placed to produce a further fourfold (4.00) magnification? (c) Is this a realistic design for a microscope?

Answer

(a) 0.501 cm

(b) Eyepiece should be 204 cm behind the objective lens.

(c) No, a distance of 205 cm (which is over 2 m) is unrealistic for the tube of a microscope.

26-30 §26.4 source ↗

An amoeba is 0.305 cm away from the 0.300 cm focal length objective lens of a microscope. (a) Where is the image formed by the objective lens? (b) What is this image’s magnification? (c) An eyepiece with a 2.00 cm focal length is placed 20.0 cm from the objective. Where is the final image? (d) What magnification is produced by the eyepiece? (e) What is the overall magnification? (See Figure 26.16.)
Figure 26.16 in the book ↗
A compound microscope composed of two lenses, an objective and an eyepiece. The objective forms a case 1 image that is larger than the object. This first image is the object for the eyepiece. The eyepiece forms a case 2 final image that is further magnified.

Answer

(a) +18.3 cm (on the eyepiece side of the objective lens)

(b) -60.0

(c) -11.3 cm (on the objective side of the eyepiece)

(d) +6.67

(e) -400

26-33 §26.5 source ↗

Unless otherwise stated, the lens-to-retina distance is 2.00 cm.

What is the angular magnification of a telescope that has a 100 cm focal length objective and a 2.50 cm focal length eyepiece?

Answer

− 40 . 0

26-35 §26.5 source ↗

Unless otherwise stated, the lens-to-retina distance is 2.00 cm.

A large reflecting telescope has an objective mirror with a 10.0 m radius of curvature. What angular magnification does it produce when a 3.00 cm focal length eyepiece is used?

Answer

− 167

26-37 §26.5 source ↗

Unless otherwise stated, the lens-to-retina distance is 2.00 cm.

A 7.5× binocular produces an angular magnification of −7.50, acting like a telescope. (Mirrors are used to make the image upright.) If the binoculars have objective lenses with a 75.0 cm focal length, what is the focal length of the eyepiece lenses?

Answer

+ 10.0 cm

26-39 §26.6 source ↗

Integrated Concepts

(a) During laser vision correction, a brief burst of 193 nm ultraviolet light is projected onto the cornea of the patient. It makes a spot 1.00 mm in diameter and deposits 0.500 mJ of energy. Calculate the depth of the layer ablated, assuming the corneal tissue has the same properties as water and is initially at 34.0ºC. The tissue’s temperature is increased to 100ºC and evaporated without further temperature increase.

(b) Does your answer imply that the shape of the cornea can be finely controlled?

Answer

(a) 0.251 μm

(b) Yes, this thickness implies that the shape of the cornea can be very finely controlled, producing normal distant vision in more than 90% of patients.

26-40 §26.6 source ↗

Critical Thinking A microscope has the objective lens 6.00 mm from an object. The focal length of the objective lens is 4.00 mm. The eyepiece has a focal length of 50.0 mm and is initially 25.0 mm from the objective lens. (a) The objective lens is moved to 5.00 mm from the objective, effectively lengthening the microscope. How does this affect the magnification of the microscope? (b) Does moving the objective increase closer to the object increase or decrease the magnification, or does it remain the same? (c) Do you think moving the objective from 6 mm to 7 mm from the object will increase or decrease the magnification of the microscope, or will it likely remain the same? (d) What indicates whether the image is upright or inverted?

Answer

(a) 1 d i + 1 d 0 = 1 f 1 d i = 1 f − 1 d 0 = 1 4.00 mm − 1 6.00 mm d i =12.0 mm m 0 =− d i d 0 =− 12.0 6.00 =−2.00 1 d ′ i + 1 d ′ 0 = 1 f 1 d ′ i = 1 f − 1 d ′ 0 = 1 50 mm − 1 13.0 mm d ′ i =−17.6 mm
m e =− −17.6 13 =1.35 m=−2(1.35)=−2.70 1 d i + 1 d 0 = 1 f 1 d i = 1 f − 1 d 0 = 1 4.00 mm − 1 5.00 mm d i =20.0 mm
m 0 =− d i d 0 =− 20.0 5.00 =−4.00 1 d ′ i + 1 d ′ 0 = 1 f 1 d ′ i = 1 f − 1 d ′ 0 = 1 50 mm − 1 5.00 mm d ′ i =−5.56 mm
m e =− −5.56 5 =1.11 m=−4(1.11)=−4.44

(b) It increased.

(c) The magnification should decrease.

(d) In both cases the image is inverted.

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Problems and answers are adapted from College Physics 2e by Paul Peter Urone, Roger Hinrichs, Kim Dirks and Manjula Sharma, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.