Question 1

Olga is nearsighted and cannot see things beyond 90 cm cm from her eyes.What is the refractive power of the contact lenses that will enable her to comfortably see distant buildings clearly?

Accepted Answer

Nearsightedness (myopia) is corrected with lenses that have a negative diopter value. The formula to find the diopter (D) needed is D = 1/f (in meters), where f is the focal length. Olga's maximum clear vision distance is 90 cm or 0.9 meters. Thus, D = 1/0.9 = 1.11, which rounds to -1.1 diopters, indicating a negative lens is needed for correction.

Question 2

A farsighted person cannot see clearly objects that are closer to her eyes than 60.0 cm.Which one of the following combinations represents the correct focal length and the refractive power of the contact lenses that will enable her to comfortably see the objects at a distance of 25.0 cm from her eyes?

Accepted Answer

Since the person is farsighted, the contact lenses have to provide diverging rays of light. The lens formula is given by:

1/f = 1/di + 1/do

where f is the focal length, di is the distance of the object from the lens, and do is the distance of the image from the lens.

Given that the person cannot see objects at a distance closer than 60.0 cm, the distance from the lens (do) to the image has to be (-25.0 cm) to bring the image closer to her eyes.

Using the lens formula and substituting the values, we get:

1/f = 1/di + 1/do
= 1/(-25.0 cm) + 1/(-42.9 cm)

Solving for f, we get:

f = +42.9 cm

Since the focal length is positive, the contact lenses have to be convex, and since the person wants to comfortably see objects at a distance of 25.0 cm, the refractive power has to be positive (to provide diverging rays of light).

The refractive power of the lenses is given by:

P = 1/f

Substituting the value of f, we get:

P = 1/(+42.9 cm)
= +2.33 diopters.

Therefore, the correct choice is (C).

Question 3

A nearsighted person wears contact lenses having a power of -0.15 diopters.What is the person's uncorrected far point?

Accepted Answer

The answer of A nearsighted person wears contact lenses having...

Question 4

A person uses a convex lens of focal length 5.0 cm as a magnifying glass.What is the angular magnification if the person's eye is relaxed and has a near point of 25 cm?

Accepted Answer

The formula for angular magnification is given by:

magnification = (1 + (25/f))/2

where f is the focal length of the lens.

Substituting f = 5.0 cm in the above formula, we get:

magnification = (1 + (25/5.0))/2 = 5.0

Therefore, the angular magnification is 5.0.

Hence, the correct answer is choice B.

Question 5

A lens of focal length 40 mm is used as a magnifier.The object being viewed is 5.3 mm long,and is positioned at the focal point of the lens.The user of the magnifier has a near point at 25 cm.What is the angular magnification of the magnifier?

Accepted Answer

The angular magnification (M) of a magnifier is given by M = 1 + (D/f), where D is the near point of the viewer (in meters) and f is the focal length of the lens (in meters). Here, D = 25 cm = 0.25 m and f = 40 mm = 0.04 m. Thus, M = 1 + (0.25/0.04) = 1 + 6.25 = 7.25. The closest answer to this calculation is 6.3, indicating a possible mistake in calculation or in the interpretation of the choices provided. However, based on the formula and the choices given, 6.3 is selected, acknowledging a discrepancy in the calculation process.

Question 6

A magnifying glass has a focal length of 7.0 cm.What is the angular magnification if the image is viewed by a person having a relaxed eye with the near point of 25 cm?

Accepted Answer

The answer of A magnifying glass has a focal length...

Question 7

A person cannot see clearly objects more than 70.0 cm away.What power of corrective lens should be prescribed if the glass is to be worn 1.00 cm in front of the eye?

Accepted Answer

The focal length of the corrective lens can be found using the formula:
1/f = 1/do + 1/di
where f is the focal length, do is the distance between the object and the lens (70 cm in this case), and di is the distance between the lens and the eye (1 cm).

Substituting the values, we get:
1/f = 1/70 + 1/1
1/f = 71/70
f = 0.98 cm

The power of the lens can be found using the formula:
P = 1/f
P = 1/0.98
P = 1.02 diopters

Since the person cannot see objects more than 70 cm away, the lens should be a concave lens, which has negative power. Therefore, the power of the corrective lens should be:
P = -1.02 diopters

The closest option is B, which is -1.45 diopters.

Question 8

A magnifying glass uses a convex lens with a refractive power of 20 diopters.What is the angular magnification if the image is to be viewed by a relaxed eye with a near point of 25 cm?

Accepted Answer

The angular magnification (M) of a magnifying glass is given by M = 1 + (D/f), where D is the near point distance (25 cm or 0.25 m) and f is the focal length of the lens. The refractive power (P) is 20 diopters, so f = 1/P = 0.05 m. Thus, M = 1 + (0.25/0.05) = 1 + 5 = 6. However, for a relaxed eye, the formula is M = D/f, so M = 0.25/0.05 = 5.0.

Question 9

A nearsighted person cannot see objects beyond 80 cm from his eyes.Which one of the following combinations represents the correct focal length and the refractive power of the contact lenses that will enable him to comfortably see the distant objects clearly?

Accepted Answer

The answer of A nearsighted person cannot see objects beyond...

Question 10

Alice is farsighted and cannot see objects clearly that are closer to her eye than 80.0 cm.What is the refractive power of the contact lenses that will enable her to comfortably see objects at a distance of 25.0 cm from her eyes?

Accepted Answer

To correct farsightedness and allow Alice to see objects at 25.0 cm, we need to calculate the lens power required to change her near point to 25.0 cm. The formula for lens power (P) in diopters is P = 1/f, where f is the focal length in meters. The lens should change the image location from 25.0 cm (0.25 m) to her comfortable viewing distance of 80.0 cm (0.80 m). Using the lens formula 1/f = 1/v - 1/u, where v is the image distance (0.80 m) and u is the object distance (-0.25 m, negative because it's on the same side of the lens as the light is coming from), we find 1/f = 1/0.80 - 1/(-0.25) = 1.25 + 4 = 5.25. Therefore, P = 5.25 diopters. Since we're dealing with a lens that corrects farsightedness, it's a converging lens, which has a positive diopter value, making the correct answer +5.25 diopters. However, given the options, it seems there might have been a calculation or typographical error in the question or answer choices. Based on the provided choices, the closest correct answer by the logic of calculation would be +4.72 diopters, but this does not match the calculation. The correct calculation approach would ideally yield a result that matches one of the given options exactly, suggesting a need to reevaluate the calculation or the provided options for accuracy.

Question 11

What is the angular magnification of a magnifying glass of focal length 4.0 cm if the image is to be viewed by a relaxed eye having a near point of 25 cm?

Accepted Answer

The answer of What is the angular magnification of a...

Question 12

The near point of a farsighted person is 100 cm.She places reading glasses very close to her eyes,and with them she can comfortably read a book at a distance of 25 cm.What is the power of the lenses in her reading glasses?

Accepted Answer

The near point of a farsighted person is farther away than normal, which means their eye has a weaker power. Using the formula 1/f = 1/di + 1/do, 
where f is the focal length of the lens, di is the image distance (25 cm), and do is the object distance (distance at which the person can see clearly without glasses, 100 cm), we can solve for the power of the lens.

1/f = 1/di + 1/do
1/f = 1/25 + 1/100
1/f = 0.04
f = 25 cm

The power of the lens is given by the formula P = 1/f, where P is in diopters.

P = 1/f 
P = 1/0.25 
P = 4

However, since the person in the question is farsighted, we need to add the additional power needed to correct the farsightedness. This can be calculated using the formula P = 1/d, where d is the distance from the lens to the person's eye (assuming the person is holding the book at arm's length, this distance is approximately 30 cm).

P = 1/d 
P = 1/0.3 
P = 3.33

Therefore, the total power of the reading glasses is 4 + 3.33 = 7.33 diopters. However, since the question asks for the power of the lenses themselves (not including the additional power needed to correct the person's farsightedness), we need to subtract the power needed to correct the farsightedness. This is 2.5 diopters (calculated using the person's near point of 100 cm).

Therefore, the power of the lenses in the reading glasses is 7.33 - 2.5 = 4.83, which is closest to 3.0 diopters (nearest among the choices).

Question 13

A lens of focal length 90 mm is used as a magnifier by a person with a near point of 25 cm.The object being viewed is 9.9 mm long,and is positioned at the focal point of the lens.What is the angle subtended by the image at infinity,in milliradians?

Accepted Answer

When an object is placed at the focal point of a lens, the image formed is at infinity, and the magnification is theoretically infinite. However, the angle subtended by the image can be calculated using the formula for angular magnification, $M = \frac{25}{f}$, where $M$ is the angular magnification, $25$ represents the near point distance in centimeters (converted to the same units as the focal length if necessary), and $f$ is the focal length of the lens. The angle subtended by the image, in radians, is then given by $\theta = \frac{h}{25}$, where $h$ is the height of the object. To find $\theta$ in milliradians (mrad), we multiply by 1000.Given that the focal length $f = 90$ mm (or 9 cm for consistency with the near point distance), and the object height $h = 9.9$ mm (or 0.99 cm), we first find the angular magnification and then calculate the angle.1. Convert the focal length to centimeters: $f = 90$ mm = 9 cm.2. Calculate the angular magnification: $M = \frac{25}{9}$.3. Calculate the angle subtended by the image at the near point distance: $\theta = \frac{h}{25} = \frac{0.99}{25}$ radians.4. Convert $\theta$ to milliradians (mrad): $\theta_{mrad} = \frac{0.99}{25} \times 1000$.However, the direct calculation above might seem misleading because when an object is at the focal point, the image formed is at infinity, and we typically consider the angular size based on the lens's ability to magnify the object as seen by the eye.The correct approach involves understanding that the angular size is related to how much the lens magnifies the object for the eye. Since the object is at the focal point, the lens creates an image that appears infinitely far away, making the object appear larger than if viewed with the naked eye at the near point distance.The angular size in milliradians can be directly calculated by considering the object size and the focal length of the lens:$\theta = \frac{h}{f} \times 1000$, where $h$ is in mm and $f$ is in mm to keep units consistent.$\theta = \frac{9.9}{90} \times 1000 = 110$ mrad.This calculation simplifies the process by directly relating the object size and the lens's focal length to find the angular size, assuming the object is placed at the focal point, creating a very large (theoretically infinite) image that subtends a significant angle to the observer's eye, calculated based on the geometry of the situation.

Question 14

Mahmood is farsighted and cannot see objects clearly that are closer to his eye than 80.0 cm.What is the focal length of the contact lenses that will enable him to comfortably see objects at a distance of 25.0 cm from his eyes?

Accepted Answer

To correct farsightedness and allow Mahmood to see objects at 25.0 cm, we use the lens formula 1/f = 1/v - 1/u, where f is the focal length of the lens, v is the image distance (considered positive for virtual images, which is the case for corrective lenses), and u is the object distance (negative if the object is in front of the lens, as per the sign convention). Mahmood wants to see objects at 25.0 cm as if they were at his far point (80.0 cm). Thus, u = -25.0 cm (since the object is in front of the lens) and v = +80.0 cm (the distance he can comfortably see without correction). Plugging these values into the lens formula gives 1/f = 1/80 - (-1/25) = 1/80 + 1/25. Solving this equation for f gives f = +36.4 cm, indicating that a lens with a focal length of +36.4 cm is required.

Question 15

When a person with a near point at 28 cm uses a certain magnifying glass,it produces an angular magnification of 7.0 with the image very distant.What is the focal length of the lens of this magnifying glass?

Accepted Answer

Firstly, we need to find the magnification produced by the magnifying glass.
Angular magnification, M = -(theta i)/(theta 0)
where theta i and theta 0 are the angles subtended by the image and object respectively, as seen by the naked eye.
For a distant object, the angle subtended by the object at the eye is approximately equal to the angle subtended by the image formed by the magnifying glass.
Therefore, in this case, M = -(theta i)/(theta)
where theta is the angle subtended by the object at the eye.
We are given that the angular magnification produced by the magnifying glass, M = 7.0 and the near point of the eye, D = 28 cm. Therefore, the distance of the object from the magnifying glass, d = D + f, where f is the focal length of the magnifying glass.
Also, for a magnifying glass, the magnification produced is given by:
M = (1 + (D/f))/(1 + (d/f))
Substituting the values of M, D and d, we get:
7.0 = (1 + (28/f))/(1 + ((28 + f)/f))
Solving for f, we get f = 4.0 cm.
Hence, the answer is C) 4.0 cm.

Question 16

A nearsighted person has a far point of 18 cm.What power contact lenses will allow this person to comfortably see distant objects clearly?

Accepted Answer

Since the person is nearsighted, they have difficulty seeing distant objects clearly. The far point (the farthest distance at which objects can be seen clearly) is 18 cm. To correct this, a concave lens is needed to diverge the light before it enters the eye. The power of the lens needed can be calculated using the formula: Power = 1/focal length in meters.

We are given the far point which is the focal length of the required lens. But the given focal length is in centimeters, so we need to convert it into meters by dividing it by 100:

Focal length = 18 cm = 0.18 m

Power = 1/0.18 = -5.56 diopters (approx.)

Therefore, the power of the contact lens needed to correct the nearsightedness and allow the person to see distant objects clearly would be -5.6 diopters.

Question 17

What power contact lens is needed to correct for farsightedness where the uncorrected near point is 75 cm?

Accepted Answer

To correct for farsightedness, we need a converging lens (positive power) that will bend light rays towards each other and bring them to a focus on the retina. The power of the lens needed can be found using the formula:

1/f = 1/distance of near point from the eye - 1/focal length of the lens

where focal length is measured in meters. If the uncorrected near point is 75 cm, then the distance of the near point from the eye is 0.75 meters. Assuming a normal far point of infinity, we can simplify the formula to:

1/f = 1/0.75 - 0

1/f = 1.33

f = 1/1.33 = 0.75 meters

Converting to diopters:

D = 1/f = 1/0.75 = 1.33 D

However, this is the power of the lens needed to bring the near point to the eye. To find the power needed for corrective lenses, we need to subtract the power of the eye's natural lens, which is about +0.5 D for a young adult.

P = 1.33 - 0.5 = +0.83 D

Therefore, we need a converging (positive) lens with a power of +0.83 D to correct for farsightedness in this case. However, this power is not one of the answer choices. The closest answer choice is A, +2.7 D, which is possible but would be a stronger correction than necessary.

Question 18

The focal length of a converging lens is 10 cm.What is the angular magnification of this lens if the image is viewed by a relaxed eye with a near point of 25 cm?

Accepted Answer

The angular magnification (M) of a lens when the image is viewed by a relaxed eye is given by the formula M = D/f, where D is the near point of the eye (25 cm in this case) and f is the focal length of the lens (10 cm). Therefore, M = 25/10 = 2.5.

Question 19

A person having a near point of 25 cm uses a magnifying glass with a focal length of 10 cm.What is the angular magnification of the glass for that person when her eyes are focused at infinity?

Accepted Answer

The angular magnification of a magnifying glass is given by the formula:

M = 1 + (d/f)

where d is the distance between the object and the magnifying glass, and f is the focal length of the magnifying glass.

In this case, the object is the eye's near point, which is 25 cm away. Therefore, d = -25 cm (negative because the object is on the same side as the eye). The focal length of the magnifying glass is given as 10 cm.

Plugging these values into the formula, we get:

M = 1 + (-25/10) = 1 - 2.5 = -1.5

The negative sign indicates that the image is inverted. However, we are only interested in the absolute value of the magnification, which is 1.5.

But since the question asks for the angular magnification, we need to use the formula:

M' = (tanφ')/(tanφ)

where φ is the angle subtended by the object at the eye, and φ' is the angle subtended by the image at the eye.

When the eye is focused at infinity, φ = 0, and φ' = -f/d = -10/-25 = 0.4