Question 1

What is the power of the lens that has a focal length of 40 cm and a diameter if 5.0 cm?

Accepted Answer

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Question 2

As a treatment for cataracts (a cloudiness of the lens of the eye), the natural lens is removed and a plastic lens is implanted. After this is done a person can see distant objects clearly, but he cannot accommodate to focus on nearby objects. If such a person wanted to read a book at a distance of 25 cm, he would have to wear contact lenses having a diopter power of

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Question 3

A person's hyperopic (farsighted)eye can focus on objects that are 75 cm away and farther. What power and what focal length contact lenses should be prescribed so this person can see objects 25 cm and farther from the eye?

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Question 4

A nearsighted person has a far point of 50 cm for his right eye and 150 cm for his left eye. For each eye, what power contact lens will allow the person to see gorgeous sunsets?

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Question 5

The power of a lens is 4.0 diopters and its diameter is 5.0 cm. What is the focal length of this lens?

Accepted Answer

The formula connecting power (P), focal length (f), and unit of measurement in diopters (D) is:
P = 1/f
where P and D are interchangeable.
Thus, 4.0 D = 1/f
f = 1/4.0 D
where D = diameter of the lens = 5.0 cm
f = 1/(4.0 x 5.0) m = 0.05 m = 5.0 x 10^-2 m = 50 mm = 5.0 x 10^1 cm
f = 25 cm (rounded to the nearest cm)

Question 6

A nearsighted man wears contact lenses of -3.00 diopters. With these lenses, his corrected near point is 25.0 cm. What is his uncorrected near point?

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Question 7

What is the focal length of the corrective contact lens needed by a nearsighted person whose far point is at 60 cm?

Accepted Answer

The focal length of the corrective lens for a nearsighted person should allow them to see distant objects clearly. Since their far point is 60 cm, the lens needs to correct vision to infinity. The formula for lens power (P) is P = 1/f, where f is the focal length in meters. To make distant objects (at infinity) appear at the person's far point (60 cm), the lens must have a focal length of -60 cm, indicating a diverging lens. This focal length will shift the image from infinity to the person's far point, correcting their nearsightedness.

Question 8

A farsighted boy has a near point at 2.3 m and requires contact lenses to correct his vision to the normal near point. What is the correct choice of lens power for the contact lenses?

Accepted Answer

The lens power required to correct farsightedness is positive. 
Near point for normal eye is 25 cm. 
Using the lens formula, 
1/f = 1/v - 1/u 
where, 
f = focal length of the lens 
v = distance of the near point for normal eye = 0.25 m 
u = distance of the near point for the farsighted eye = 2.3 m

Substituting the values, 
1/f = 1/0.25 - 1/2.3 
1/f = 4.24 
f = 0.236 m or 236 mm

Hence, the lens power required is 1/f = 4.24 diopters. 
Since the boy requires correction to the normal near point, the lens power required is +4.24 D. 
However, the answer choices given are in decimals. Rounding off +4.24 to one decimal place gives +4.2. 
To find the power of contact lens required, we use the formula: 
Power of contact lens = Power required - Power of glasses/contact lenses already worn 
Since the boy requires contact lenses and not glasses, we assume that he is not wearing any glasses currently. Therefore, the power of the contact lens required is +4.2 D (closest matching answer choice is D).

Question 9

The near point of a farsighted person is 65 cm. What power contact lens must he use to correct this problem and be able to read a book at a normal near point of 25 cm?

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Question 10

What power contact lens must be used to correct the vision of a nearsighted person whose far point is 40 cm?

Accepted Answer

The power of the lens needed to correct nearsightedness is negative. To find the power, we use the formula P = -1/f, where P is the power in diopters, and f is the focal length of the lens in meters. The far point of a nearsighted person is the distance at which they can see objects clearly, and it is less than infinity. Therefore, we convert the far point to meters by dividing it by 100 (40 cm = 0.4 m) and use it as the focal length. P = -1/0.4m = -2.5 diopters, which is option B.

Question 11

A myopic (nearsighted)child wears contact lenses that allow her to have clear distant vision. The power of the lenses of her eyeglasses is -3.00 diopters. Without the corrective lenses, what is the far point of the girl?

Accepted Answer

The far point of a myopic person can be calculated using the formula: Far point = 1/Power of the lens in diopters. Therefore, for a lens with a power of -3.00 diopters, the far point = 1/3 = 0.33 meters.

Question 12

A nearsighted person has a far point that is 4.2 m from his eyes. What power contact lenses must he wear to allow him to focus on distant mountains?

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Question 13

What is the uncorrected near point of a farsighted person who can comfortably read a newspaper held 25 cm from his eyes when he wears +3.33-diopter contact lenses?

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Question 14

What power contact lens should be used to correct the vision of a farsighted person whose near point is 80 cm so she can see things clearly that are 25 cm in front of her?

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Question 15

A machinist with normal vision has a near point at 25 cm. This machinist wears +4.25-diopter eyeglasses in order to do very close work. With these eyeglasses, what is the near point of the machinist? Assume that he wears the glasses extremely close to his eyes.

Accepted Answer

The near point with the glasses can be found using the formula: 
1/f = 1/d_o + 1/d_i, where f is the power of the lens (in meters), d_o is the object distance (25 cm), and d_i is the image distance (what we want to find).
Solving for d_i: 
1/4.25 = 1/0.25 + 1/d_i
d_i = 0.12 m = 12 cm.

Question 16

A nearsighted person has a far point that is 4.2 m from his eyes. What focal length lenses must he use in his contact lenses to allow him to focus on very distant mountains?

Accepted Answer

The correct focal length for the lenses is -4.2 m. This is because a nearsighted person needs diverging lenses to correct their vision, which have a negative focal length. The far point distance indicates the maximum distance at which the person can see clearly without correction. To see distant objects clearly, the lenses must have a focal length that effectively extends this far point to infinity, which is achieved with a focal length that matches the negative of the far point distance.

Question 17

What is the focal length of a contact lens that will allow a person with a near point of 125 cm to read a physics book held 25.0 cm from his eyes?

Accepted Answer

Using the lens formula,

1/f = 1/do + 1/di

where f is the focal length, do is the object distance (125 cm in this case), and di is the image distance (25 cm in this case).

We can solve for f:

1/f = 1/125 + 1/25
1/f = 0.008
f = 125

So the focal length of the contact lens should be 31.3 cm (since the lens is closest to the eye, we use a positive value for f).

Therefore, the best choice is E.

Question 18

A farsighted girl has a near point at 2.0 m but has forgotten her glasses at home. The girl borrows eyeglasses that have a power of +2.75 diopters. With these eyeglasses, what is the near point of the girl, assuming that she wears them extremely close to her eyes?

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Question 19

Without glasses, a person can comfortably read a magazine when it is held at 60 cm from his eyes. What should be the focal length of his contact lenses to allow him to read the magazine comfortably at a distance of 30 cm?

Accepted Answer

The formula for calculating power of a lens is:
P = 1/f
where P is power of the lens in dioptres and f is the focal length of the lens in metres.

In this problem, we need to find the focal length of the contact lenses that will enable the person to read a magazine comfortably at a distance of 30 cm. 
Let the required focal length be f1. 
Using the formula, 
P = 1/f1
or f1 = 1/P

We can find P from the given information. 
When the person reads the magazine comfortably at 60 cm, 
the image of the magazine is formed at the person's near point. 
The near point of a normal eye is about 25 cm. 
So, the distance between the eye and the image of the magazine is 60 - 25 = 35 cm. 
Using the formula, 
1/f = 1/v - 1/u 
where v is the distance of the image from the eye, u is the distance of the object from the eye,
we get 
1/f = 1/35 - 1/60 
Solving this equation gives f = +84 cm 
(We take the positive value because the image is formed on the same side as the object and the lens required will be a convex lens which has a positive focal length.)
Now using the formula, 
P = 1/f1 
gives 
f1 = 1/P = 1/0.84 = +1.19 cm 
Therefore, the required focal length of the contact lenses is +1.19 cm or approximately +1.2 cm. 
This means the lenses should have a power of about 83.3 dioptres (because power and focal length are reciprocal).

So, the correct choice is D) +60 cm.

Question 20

What is the power of a lens that has a focal length of -40 cm and a width of 2.0 cm?

Accepted Answer

The power of a lens (P) is given by the formula P = 1/f, where f is the focal length in meters. Converting -40 cm to meters gives -0.4 m. Therefore, P = 1/(-0.4) = -2.5 diopters.

Quiz 19: Optical Instruments