Question 1

A concave mirror has a radius of 2.0 m. What is the refractive power of a lens of the same focal length as the mirror?

Accepted Answer

The focal length of a concave mirror is half of its radius, so the focal length of this mirror is 1.0 m. The refractive power of a lens is given by the formula P = 1/f, where P is the power in diopters and f is the focal length in meters. Therefore, the power of a lens with the same focal length as the mirror is P = 1/1.0 m = 1.0 D.

Question 2

A lens of focal length 10 cm is placed in contact with a lens of focal length −15 cm. What is the focal length of the combination?

Accepted Answer

The answer of A lens of focal length 10 cm...

Question 3

For maximum angular magnification in a compound microscope, the objective focal length (f_o) and the eyepiece focal length (f_e) should be chosen so that

Accepted Answer

For maximum angular magnification in a compound microscope, both the objective and the eyepiece should have the smallest possible focal lengths. This configuration increases the magnification factor of the microscope.

Question 4

What magnifying power is attained using a compound microscope with the following specifications: objective focal length 1.50 cm, eyepiece focal length 2.42 cm, tube length 16.0 cm? Assume a near point of 25.0 cm and that the microscope is adjusted for viewing an image at infinity.

Accepted Answer

The magnifying power of a compound microscope is given by the formula:

M = (25 cm)/(f0/fe)(1+25/cm)

where f0 is the objective focal length, fe is the eyepiece focal length, and 25 cm is the near point.

Plugging in the values given, we get:

M = (25 cm)/[(1.50 cm)/(2.42 cm)](1+25/16) = 110

Therefore, the magnifying power attained is 110, which corresponds to choice letter A.

Question 5

A camera with interchangeable lenses produces an image of a distant object that is 3.0 mm wide on the film when the 50.0 mm lens is used. When the 135 mm lens is used, how wide is the image of the object on the film?

Accepted Answer

The size of the image on the film is directly proportional to the focal length of the lens used. The ratio of the focal lengths (135 mm / 50 mm = 2.7) can be used to find the new image width: 3.0 mm * 2.7 = 8.1 mm.

Question 6

To get the same exposure on the film when the aperture diameter of a camera lens is doubled, what change should be made in the shutter speed?

Accepted Answer

Doubling the aperture diameter increases the area by a factor of four, allowing four times more light to hit the film. To maintain the same exposure, the shutter speed should be quartered, reducing the exposure time by a factor of four.

Question 7

A farsighted person cannot see clearly closer than 2.0 m. What power contact lenses would correct this near point to 25 cm?

Accepted Answer

The near point of a farsighted person is farther away than normal. To correct for this problem, we need to add converging power (positive diopters) to the eye. The power of the contact lenses required to correct the near point to 25 cm can be found using the formula:
1/f = 1/do + 1/di
where f is the focal length of the lens, do is the object distance, and di is the image distance. In this case, the object distance is 25 cm (the desired near point), and the image distance is 200 cm (the uncorrected near point). Substituting these values into the formula, we get:
1/f = 1/200 - 1/25
Solving for f, we get:
f = 3.5 cm
Therefore, the power of the contact lenses needed to correct for this farsightedness is:
P = 1/f = 1/0.035 m = 28.6 D
However, contact lenses are much closer to the eye than glasses, so we need to account for the distance between the contact lens and the eye. If we assume a distance of 1 cm between the contact lens and the cornea, we can use the formula:
P' = P - 1/d
where P is the calculated power of the contact lens, and d is the distance between the contact lens and the cornea. Substituting the values, we get:
P' = 28.6 D - 1/0.01 m = 18.6 D
Converting from diopters to meters, we get:
P' = 1/f' = 1/0.0186 m = 53.8 D
Finally, since contact lenses are available in 0.25 D increments, the closest match is 54 D, which is equivalent to 3.7 D. Therefore, the correct answer is D) 3.5 D.

Question 8

A camera with a 50.0 mm lens is designed to focus on objects at distances ranging from 1.00 m to infinity. How far must the lens be able to move, relative to the image sensor plane, to focus over this range?

Accepted Answer

To find the range of movement, use the lens formula 1/f = 1/v - 1/u. For u = infinity, v = f = 50 mm. For u = 1 m, solve for v: 1/50 = 1/v - 1/1000, giving v ≈ 52.63 mm. The lens must move 52.63 mm - 50 mm = 2.63 mm.

Question 9

A 5.0 cm focal length lens is used as a magnifier. Assuming a near point of 25 cm, about what magnification is attained?

Accepted Answer

The magnification (M) of a magnifying lens is given by the formula M = 1 + (D/f), where D is the near point distance (25 cm) and f is the focal length (5 cm). Substituting the values, M = 1 + (25/5) = 1 + 5 = 6. However, the closest option to this calculation is 5.

Question 10

Increasing the aperture size on a camera

Accepted Answer

Increasing the aperture size on a camera will reduce the depth of field, meaning that less of the scene will be in focus. This requires a shorter exposure time to maintain proper exposure because more light is entering the camera.

Question 11

A nearsighted person wears contact lenses of refractive power −6.0 D. What is his far point?

Accepted Answer

The far point of a nearsighted person is the distance at which the person can see objects clearly without accommodation. The refractive power of the lens needed to correct for this is equal in magnitude but opposite in sign to the person's refractive error. Therefore, the far point for this person is at a distance of 1 meter (100 cm) away from the eye, which is equal to the reciprocal of the refractive power in diopters. 
Far point = 1/(-6.0 D) = 16.7 cm ≈ 17 cm (rounded to nearest integer)

Question 12

Three lenses, each of focal length 90 cm, are placed in contact. What is the focal length of the combination?

Accepted Answer

When lenses are in contact, the effective focal length is given by:
1/f = 1/f1 + 1/f2 + 1/f3 +...

Since all three lenses have the same focal length of 90 cm, the equation simplifies to:
1/f = 1/90 + 1/90 + 1/90
1/f = 3/90
f = 30 cm

Therefore, the combined focal length of the three lenses is 30 cm, which corresponds to answer choice E.

Question 13

The loss of accommodation that comes with aging is called

Accepted Answer

Presbyopia is the age-related loss of accommodation, or the ability to focus on close objects. Myopia is nearsightedness, hyperopia is farsightedness, retinalopia and macularopia are not recognized terms in optometry.

Question 14

The refractive power of a lens of focal length 40 cm is

Accepted Answer

The refractive power of a lens is given by the formula P = 1/f, where f is the focal length of the lens. Substituting f = 40 cm gives P = 1/40 cm = 0.025 cm⁻¹ = 2.5 D. Therefore, the correct answer is choice C.

Question 15

A person has his nearsightedness corrected by lenses of refractive power −5.00 D. Without his glasses, his near point was 12.5 cm. With his glasses on, what is his near point? Assume the lenses are very close to his eyes.

Accepted Answer

The near point of a person with normal vision is 25 cm. The formula for the lens' corrective power is $P = \frac{1}{f}$, where $P$ is the power in diopters and $f$ is the focal length in meters. For a lens with a power of -5.00 D, the focal length is $f = \frac{1}{P} = -0.20$ m or -20 cm. The negative sign indicates that it's a diverging lens, suitable for correcting nearsightedness. The formula for the image distance ($v$) using the lens formula is $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$, where $u$ is the object distance (in this case, the near point without glasses, -12.5 cm, taking the sign convention into account). Rearranging for $v$, we get $v = \frac{uf}{u - f}$. Substituting $u = -12.5$ cm and $f = -20$ cm, we find $v = \frac{(-12.5)(-20)}{-12.5 + 20} = \frac{250}{7.5} = 33.3$ cm. This means, with glasses, the person's near point is effectively 33.3 cm from his eyes, allowing him to see objects clearly at a distance where he couldn't without glasses.

Question 16

A small object has an angular diameter of 0.50°. When viewed through a magnifier, it has an angular diameter of 8.0°. What is the angular magnification in this case?

Accepted Answer

Angular magnification = (Angular diameter of image)/(Angular diameter of object)= 8.0°/0.50° = 16

Question 17

A lens has a refractive power of −1.50 D. What is its focal length?

Accepted Answer

The focal length $f$ in meters is given by the reciprocal of the lens power in diopters (D), so $f = \frac{1}{-1.50} = -0.666...$ meters, which is -66.7 cm.

Question 18

What kind of images do the two lenses in a compound microscope form?

Accepted Answer

The lens closest to the object in a compound microscope (the objective lens) forms a real, inverted image. This real image is then magnified by the second lens (the eyepiece), which forms a virtual image that can be observed by the eye.

Question 19

A nearsighted person has a far point of 33 cm. What focal length contact lens would allow her to see in the distance?

Accepted Answer

For a nearsighted person to see distant objects clearly, the far point of their vision must be corrected to infinity. The lens formula 1/f = 1/v - 1/u can be used, where f is the focal length of the lens, v is the image distance (considered infinity for distant objects), and u is the object distance (the far point of the person, which is -33 cm, negative because it is on the same side of the lens as the light is coming from). Solving for f gives us -33 cm, indicating a diverging lens is needed with a focal length of -33 cm.

Question 20

The eye focuses by

Accepted Answer

The eye focuses by changing the focal length of the lens. When the ciliary muscles surrounding the lens contract, the lens becomes thicker, increasing its refractive power and allowing the eye to focus on near objects. When the ciliary muscles relax, the lens becomes flatter, decreasing its refractive power and allowing the eye to focus on far objects. This process is known as accommodation. The other options (A, B, C, E) are not correct because they do not accurately describe how the eye focuses.

Quiz 24: Optical Instruments