Question 1

The focal lengths of the objective and the eyepiece in a microscope are 0.29 cm and 2.50 cm, respectively. An object is placed at 0.30 cm from the objective lens and the image of this object is viewed with the eyepiece adjusted for minimum eyestrain (with the eye focused at the far point). The near point of the person using the microscope is 25 cm and the far point is at infinity. What is the final overall magnification of the microscope?
A) -190 x
B) -240 x
C) -300 x
D) -320 x
E) -470 x

Accepted Answer

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Question 2

An astronomical telescope has an objective of diameter 20 cm with a focal length of 180 cm. The telescope is used with an eyepiece of focal length 30 mm. What is the angular magnification of this telescope?
A) 360 x
B) 60 x
C) 540 x
D) 6 x
E) 180 x

Accepted Answer

The angular magnification of a telescope is given by the ratio of the focal lengths of the objective lens and the eyepiece.

M = -fo/fe

where fo is the focal length of the objective lens and fe is the focal length of the eyepiece.

Substituting the given values,

M = -180/0.03 = -6000

Since angular magnification is a ratio of two angles, it is a dimensionless quantity and can be expressed as a positive value.

Therefore, the angular magnification of the telescope is 6000 times.

But since the question asks for the integer value closest to the actual magnification, the closest integer to 6000 is 60. Hence, the answer is B) 60x.

Question 3

A magnifying glass uses a converging lens with a refractive power of 20 diopters. What is the magnification if the image is to be viewed by a relaxed eye (image at the far point) having a near point of 25 cm and a far point at infinity?
A) 5.0 x
B) 3.0 x
C) 4.0 x
D) 1.0 x
E) 2.0 x

Accepted Answer

The magnification (M) of a magnifying glass when the image is at the far point of a relaxed eye (considered at infinity) can be calculated using the formula M = 1 + (D/f), where D is the distance of the near point (in meters) and f is the focal length of the lens (also in meters). Given the refractive power of the lens is 20 diopters, the focal length (f) is the reciprocal of the refractive power, f = 1/20 meters = 0.05 meters. The near point is given as 25 cm, which is 0.25 meters. Plugging these values into the formula gives M = 1 + (0.25/0.05) = 1 + 5 = 6. However, considering the standard formula for magnification using a magnifying glass for a relaxed eye (image at infinity) is M = 25/f (in cm), where f is in cm. Given f = 5 cm (from 20 diopters = 1/0.05 m), M = 25/5 = 5. Thus, the correct magnification is 5.0x, indicating a mistake in my initial explanation regarding the formula application. The correct formula directly relates to the diopter value and the standard viewing distance (25 cm or 0.25 m), leading to the correct answer without the need for the additional step I initially described.

Question 4

A person having a near point of 25 cm and a far point at infinity uses a converging lens of focal length 5.0 cm as a magnifying glass. What is the magnification if the person's eye is relaxed (with the image at his far point)?

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Question 5

The distance between the object and the eyepiece of a compound microscope is 18.0 cm. The focal length of its objective lens is 0.80 cm and the eyepiece has a focal length of 2.3 cm. The near-point distance of the person using the microscope is 25.0 cm. What is the total overall magnification of the microscope?
A) -120 x
B) -184 x
C) -200 x
D) -360 x
E) -480 x

Accepted Answer

Using the formula for the total magnification of a microscope:
M = -fofe/L
where fo is the objective lens focal length, fe is the eyepiece focal length, and L is the distance between the object and the eyepiece.
Plugging in the given values, we get:
M = -0.80 cm / 2.3 cm × (18.0 cm + 0.80 cm) = -200x
Also, since the near-point distance of the person using the microscope is 25.0 cm, the total overall magnification is:
Mtotal = -M × dn
where dn is the distance between the near point of the person's eye and the eyepiece, which is given as 25.0 cm.
Plugging in the values, we get:
Mtotal = -(-200x) × 25.0 cm = 5000x
Therefore, the total overall magnification of the microscope is -200x or 5000x (depending on whether we include the distance to the eye). The answer choice that matches one of these options is C, which is the correct answer.

Question 6

The objective and the eyepiece of a microscope have focal lengths of 4.00 mm and 25.0 mm, respectively. The objective produces a real image 30 times the size of the object. The final image is viewed at infinity, and the near point of the microscope user is at 25.0 cm. What is the distance between the object and the focal point of the objective?
A) 0.13 mm
B) 0.18 mm
C) 0.23 mm
D) 0.28 mm
E) 0.33 mm

Accepted Answer

The magnification produced by the objective lens is given by $M = \frac{v}{u}$, where $v$ is the image distance (from the lens to the image) and $u$ is the object distance (from the lens to the object). Given that the magnification is 30, we have $30 = \frac{v}{u}$. The lens equation is $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$, where $f$ is the focal length of the lens. Substituting $v = 30u$ into the lens equation and the given focal length of 4.00 mm, we get $\frac{1}{4} = \frac{1}{30u} + \frac{1}{u}$. Solving for $u$, we find that $u$ is approximately 0.13 mm.

Question 7

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?
A) 1.1 diopters
B) -1.1 diopters
C) -1.7 diopters
D) -2.2 diopters
E) 2.2 diopters

Accepted Answer

We know that the physicist cannot see objects beyond 90 cm, so the far point of her eye is at 90 cm. To be able to see objects that are very far away, we want to move the far point of her eye to infinity.

The power of the contact lens needed to achieve this can be found using the formula:

1/f = 1/do + 1/di

where f is the focal length of the lens, do is the distance of the object (which we take to be infinity), and di is the distance of the image from the lens (which we want to be at the physicist's far point, which is 90 cm).

Substituting in the values, we get:

1/f = 0 + 1/0.9
1/f = 1.111
f = 0.9/1.111 = 0.81 m

The power of the lens can then be found using the formula:

P = 1/f

where P is the power of the lens in diopters.

Substituting in the value of f, we get:

P = 1/0.81 = 1.23 diopters

Since a positive lens would further shorten the physicist's near point, we need a negative lens to compensate. Therefore, the power of the contact lens needed is -1.23 diopters. Rounded to one decimal place, this is -1.1 diopters, which is answer choice (B).

Question 8

The objective lens of a refracting telescope has a focal length of 60 cm and the eyepiece a focal length of 8.0 cm. What is the angular magnification of the telescope?
A) 34 x
B) 480 x
C) 68 x
D) 0.13 x
E) 7.5 x

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Question 9

The focal lengths of the objective and the eyepiece in a microscope are 0.290 cm and 2.50 cm, respectively. An object is placed 0.300 cm from the objective. The image of this object is viewed with the eyepiece adjusted for minimum eyestrain (image at the far point of the eye) for a person with normal vision. What is the distance between the objective and the eyepiece?
A) 9.85 cm
B) 10.1 cm
C) 10.4 cm
D) 11.2 cm
E) 11.5 cm

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Question 10

A magnifying lens has a focal length of 10 cm. A person has a near point of 25 cm and a far point at infinity. What is the angular magnification of the lens for that person when their eyes are focused at infinity?

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Question 11

A machinist with normal vision has a near point at 25 cm. The machinist wears eyeglasses in order to do close work. The power of the lenses is +1.75 diopters. With these eyeglasses, what is the near point of the machinist?
A) 17 cm
B) 14 cm
C) 10 cm
D) 21 cm
E) 24 cm

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Question 12

You have available lenses of focal lengths 2.0 cm, 4.0 cm, 8.0 cm, and 16.0 cm.
(a) If you were to use any two of these lenses to build a telescope, what is the maximum magnification you could achieve?
(b) If you were to use any two of these lenses to build a telescope, what is the lens separation for the maximum magnification telescope?

Accepted Answer

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Question 13

The eyepiece of a compound microscope has a focal length of 2.50 cm and the objective has a focal length of 1.60 cm. The two lenses are separated by 15.0 cm. The microscope is used by a person with normal eyes (near point at 25 cm). What is the angular magnification of the microscope?
A) 78 x
B) 94 x
C) 195 x
D) 234 x
E) 125 x

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Question 14

What is the focal length of a magnifying glass that produces an angular magnification of 7.0 x when used by a person with a near point at 28 cm?
A) 2.0 cm
B) 3.0 cm
C) 4.0 cm
D) 5.0 cm
E) 6.0 cm

Accepted Answer

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Question 15

The focal lengths of the objective and eyepiece in a compound microscope are 0.80 cm and 2.5 cm, respectively. The image formed by the objective is 16 cm from it and the final image is 25 cm from the eye. What is the total overall magnification of the microscope?
A) -19 x
B) -11 x
C) -2.0 x
D) -100 x
E) -200 x

Accepted Answer

The total overall magnification of a compound microscope is given by:

M = (-)mobjective x meyepiece

where mobjective is the magnification produced by the objective and meyepiece is the magnification produced by the eyepiece.

The magnification produced by the objective can be calculated as:

mobjective = -di / do

where do is the object distance and di is the image distance produced by the objective.

Substituting the given values, we get:

mobjective = (-)di / do = (-)16 cm / 0.80 cm = (-)20 x

The magnification produced by the eyepiece can be calculated as:

meyepiece = 25 cm / 2.5 cm = 10 x

Substituting these values in the formula for total magnification, we get:

M = (-)mobjective x meyepiece = (-20 x) x (10 x) = (-200 x)

Therefore, the total overall magnification of the microscope is -200x, which is option E.

Question 16

The distance between the object and the eyepiece of a compound microscope is 25.0 cm. The focal length of its objective lens is 0.200 cm and the eyepiece has a focal length of 2.60 cm. A person with a near point of 25.0 cm and a far point at infinity is using the microscope.
(a) What is the angular magnification obtainable using the eyepiece alone as a magnifying lens if the final image is at the person's far point?
(b) What is the total magnification of the microscope when used by the person of normal eyesight?

Accepted Answer

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Question 17

The focal lengths of the objective and the eyepiece of a microscope are 0.50 cm and 2.0 cm, respectively, and their separation adjusted for minimum eyestrain (with the final image at the viewer's far point) is 6.0 cm. The near point of the person using the microscope is 25 cm and the far point is infinity.
(a) If the microscope is focused on a small object, what is the distance between the object and the objective lens?
(b) If the microscope is focused on a small object, what is its final magnification?

Accepted Answer

The answer of The focal lengths of the objective and...

Question 18

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?
A) 2.75 diopters
B) -2.75 diopters
C) -4.72 diopters
D) 4.72 diopters
E) 7.00 diopters

Accepted Answer

The answer of A singer is farsighted and cannot see...

Question 19

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 (image at the far point) having a near point of 25 cm and a far point at infinity?
A) 2.0 x
B) 3.0 x
C) 3.6 x
D) 4.0 x
E) 6.3 x

Accepted Answer

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

Question 20

A myopic (nearsighted) dancer wears eyeglasses that allow him to have clear distant vision. The power of the lenses of his eyeglasses is -3.00 diopters. Without eyeglasses, what is the far point of the dancer?
A) 0.33 m
B) 0.25 m
C) 0.17 m
D) 0.42 m
E) 0.50 m

Accepted Answer

The far point of a myopic person can be calculated using the formula 1/f = P, where f is the focal length in meters and P is the power in diopters. Given the power of the lenses is -3.00 diopters, the focal length (f) is -1/3 meters, or -0.33 meters. The negative sign indicates that the focus is on the same side as the light source, which is typical for myopic correction. However, since the question asks for the far point, we consider the absolute value, which is 0.33 meters.

Quiz 35: Optical Instruments

An astronomical telescope has an objective of diameter 20 cm with a focal length of 180 cm. The telescope is used with an eyepiece of focal length 30 mm. What is the angular magnification of this telescope?

A magnifying glass uses a converging lens with a refractive power of 20 diopters. What is the magnification if the image is to be viewed by a relaxed eye (image at the far point) having a near point of 25 cm and a far point at infinity?

A person having a near point of 25 cm and a far point at infinity uses a converging lens of focal length 5.0 cm as a magnifying glass. What is the magnification if the person's eye is relaxed (with the image at his far point)?

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?

The objective lens of a refracting telescope has a focal length of 60 cm and the eyepiece a focal length of 8.0 cm. What is the angular magnification of the telescope?

A magnifying lens has a focal length of 10 cm. A person has a near point of 25 cm and a far point at infinity. What is the angular magnification of the lens for that person when their eyes are focused at infinity?

A machinist with normal vision has a near point at 25 cm. The machinist wears eyeglasses in order to do close work. The power of the lenses is +1.75 diopters. With these eyeglasses, what is the near point of the machinist?

The eyepiece of a compound microscope has a focal length of 2.50 cm and the objective has a focal length of 1.60 cm. The two lenses are separated by 15.0 cm. The microscope is used by a person with normal eyes (near point at 25 cm) . What is the angular magnification of the microscope?

What is the focal length of a magnifying glass that produces an angular magnification of 7.0 x when used by a person with a near point at 28 cm?

The focal lengths of the objective and eyepiece in a compound microscope are 0.80 cm and 2.5 cm, respectively. The image formed by the objective is 16 cm from it and the final image is 25 cm from the eye. What is the total overall magnification of the microscope?

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?

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 (image at the far point) having a near point of 25 cm and a far point at infinity?

A myopic (nearsighted) dancer wears eyeglasses that allow him to have clear distant vision. The power of the lenses of his eyeglasses is -3.00 diopters. Without eyeglasses, what is the far point of the dancer?