Question 1

A thin double convex lens is to focus the image of an object onto a screen so that the image is life-sized. The lens surfaces have equal radii of curvature of 112 cm, and the refractive index of the lens material is 1.500.
(a) How far from the lens should the screen be placed?
(b) How far is the screen from the object?

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Question 2

A 35-mm camera equipped with a 95 mm focal length lens is used to photograph a tree that is 8.0 m tall. A 32 mm high image of the tree on the film is needed. The required distance, between the tree and the camera lens, to take the photograph is closest to

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Question 3

A person can read the newspaper when it is held at 60 cm from his eyes. What should the focal length of his contact lenses be to allow him to read the newspaper comfortably at a distance of 30 cm?

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Question 4

A double-concave lens has equal radii of curvature of 15.1 cm. An object placed 14.2 cm from the lens forms a virtual image 5.29 cm from the lens. What is the index of refraction of the lens material?

Accepted Answer

The lens equation is 1/f = 1/do + 1/di, where f is the focal length, do is the object distance, and di is the image distance. For a double-concave lens, f is negative. We can use the thin lens equation to find the focal length: 1/f = 1/do + 1/di = 1/14.2 cm + 1/(-5.29 cm) = -0.0423 cm^-1. Therefore, f = -23.6 cm. The lens maker's equation is 1/f = (n-1)(1/R1 + 1/R2), where n is the index of refraction of the lens material, and R1 and R2 are the radii of curvature of the lens surfaces. For a double-concave lens, R1 = R2 = -15.1 cm. Substituting these values into the lens maker's equation, we get -0.0423 cm^-1 = (n-1)(1/-15.1 cm + 1/-15.1 cm). Solving for n, we get n = 1.90.

Question 5

A camera lens has a focal length of 50.0 mm and an aperture setting of f/4.00. What is the aperture diameter of this lens?

Accepted Answer

The aperture diameter is calculated by dividing the focal length by the f-number. So, 50.0 mm / 4.00 = 12.5 mm.

Question 6

What is the focal length of the corrective contact lens needed by a nearsighted person whose far point is 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) in diopters is P = 1/f (in meters), where f is the focal length. To make distant objects (at infinity) appear at the person's far point (60 cm or 0.6 meters), the lens must have a focal length of -0.6 meters or -60 cm. Negative because it's a diverging lens, suitable for nearsightedness correction.

Question 7

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 the vision of a nearsighted person can be found using the formula: P = -1/f, where P is the power of the lens in diopters and f is the focal length in meters. 
The far point of a nearsighted person is the maximum distance at which they can see objects clearly, which is less than infinity. In this case, the far point is given as 40 cm or 0.4 m. 
Using the formula, P = -1/f, we get P = -1/0.4 = -2.5 diopters. Therefore, the correct choice is B, which is -2.5 diopters.

Question 8

A man is nearsighted and cannot see things clearly beyond 110from his eyes. What is the focal length of the contact lenses that will enable him to see very distant objects clearly?

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Question 9

A compound lens is made by joining the plane surfaces of two thin plano-convex lenses of different glasses. The radius of curvature of each convex surface is 80 cm. The indices of refraction of the two glasses are 1.50 and 1.60. What is the focal length of the compound lens?

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Question 10

A singer is farsighted and cannot see objects clearly that are closer than 80.0 cm from his unaided eye. What is the refractive power of the contact lenses that will move his near point to a distance of 25.0 cm from his eye?

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Question 11

The near point of a person's uncorrected eyes is 125.0 cm and the far point is at infinity. What is the focal length of a contact lens that will move the near point to 25.0 cm from this person's eyes?

Accepted Answer

The formula for determining the focal length of a contact lens is:

1/f = 1/far point - 1/near point

We are given that the far point is at infinity, so we can simplify this to:

1/f = 0 - 1/125

Solving for f, we get:

f = -125

This means that a contact lens with a focal length of -125 cm would be needed to bring the near point to 125 cm. However, we want the near point to be 25 cm, so we need to adjust our formula:

1/f = 0 - 1/25

Solving for f, we get:

f = -25

So the focal length of the contact lens that will move the near point to 25 cm is -25 cm, or choice E.

Question 12

A nearsighted physicist cannot see things clearly beyond 90 cm from her eyes. What is the power of the contact lenses that will enable her to see very distant objects clearly?

Accepted Answer

To calculate the power of the required contact lens, we use the following formula:
Power of lens (in diopters) = 1/focal length (in meters)
Here, the physicist cannot see objects beyond 90 cm, which is 0.9 meters. So, the focal length required for the contact lens should be:
focal length = 0.9 meters
Substituting this value in the formula above, we get:
Power of lens = 1/0.9 = 1.11 diopters
However, since the physicist is nearsighted, the lens should be a concave lens (or a diverging lens), which has a negative power. So, the power of the required contact lens is:
Power of lens = -1.1 diopters (closest option is B))

Question 13

A nearsighted person has her distant vision corrected using a -2.0-diopter contact lens. Her uncorrected near point is 15 cm. What is her near point using this lens if the lens is 2.0 cm from the eye?

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Question 14

A person's eye lens is 2.7 cm away from the retina. This lens has a near point of 25 cm and a far point at infinity.
(a) What must the focal length of this lens be in order for an object placed at the near point of the eye to focus on the retina?
(b) What must the focal length of this lens be in order for an object placed at the far point of the eye to focus on the retina?

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Question 15

In a 35.0-mm single lens reflex camera (SLR), the distance from the lens to the film is varied in order to focus on objects at varying distances. Over what distance range must a lens of 55-mm focal length vary if the camera is to be able to focus on objects ranging in distance from infinity down to 0.60 m from the camera?

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Question 16

A double-convex thin lens is made of glass with an index of refraction of 1.52. The radii of curvature of the faces of the lens are 60 cm and 72 cm. What is the focal length of the lens?

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Question 17

What is the uncorrected near point of a person who has a near point that is 25 cm from his eyes when he is wearing 3.33-diopter contact lenses?

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Question 18

The near point of a farsighted person's uncorrected eyes is 80 cm. What power contact lens should be used to move the near point to 25 cm from this person's eyes?

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Question 19

The focal length of a thin lens is 40 mm and its aperture diameter is 10 mm. What is the f-number of this lens?

Accepted Answer

The f-number (N) of a lens is defined as the ratio of the focal length (f) to the diameter of the aperture (D), so N = f/D. Given a focal length of 40 mm and an aperture diameter of 10 mm, the f-number is 40 mm / 10 mm = 4.0, which is represented as f/4.0.

Question 20

A lens is made with a focal length of -40 cm using a material with index of refraction 1.50. A second lens is made with the SAME GEOMETRY as the first lens, but using a material having refractive index of 2.00. What is the focal length of the second lens?

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Quiz 33: Geometric Optics