Question 1

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 2

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?

Accepted Answer

The angular magnification is given by:
$$M= \frac{	heta'}{	heta} = \frac{1}{1-\frac{d}{f}}$$
where $	heta'$ is the angle subtented by the image, $	heta$ is the angle subtended by the object, $d$ is the distance of the object from the lens, and $f$ is the focal length of the lens.

The near point of a person is 28 cm, which means the person can only see objects clearly when they are at a distance of 28 cm or closer. In this case, we take the object distance to be equal to the near point $d=28	ext{ cm}$. The angular magnification is given to be $M=7.0$. Thus, we have:
$$7.0 = \frac{	heta'}{	heta} = \frac{1}{1-\frac{28	ext{ cm}}{f}}$$
Solving for $f$, we get:
$$f=4.0	ext{ cm}$$

Therefore, the best choice is C) 4.0 cm.

Question 3

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?

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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?

Accepted Answer

We can calculate the magnification of the objective lens with the formula M(o) = (-d(o)/f(o)) = (-18.0 cm/0.80 cm) = -22.5 x
We can calculate the magnification of the eyepiece with the formula M(e) = (25.0 cm/f(e)) = (25.0 cm/2.3 cm) = 10.87 x (Note: since the eyepiece is a convex lens, the overall magnification is positive)
Using the formula for total magnification M = M(o) x M(e), we get M = (-22.5 x) x (10.87 x) = -244.4 x, which can be rounded to -200 x. Therefore, the best choice is C.

Question 6

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?

Accepted Answer

The angular magnification of a telescope is given by the ratio of the focal length of the objective to the focal length of the eyepiece. 
So, the angular magnification of this telescope is:
M = focal length of objective/focal length of eyepiece
M = (180 cm)/(0.3 cm)
M = 600
But, this is the linear magnification. To find the angular magnification, we use the formula:
Angular magnification = M*(wavelength of light)/(diameter of objective)
Assuming light of wavelength 550 nm, we get:
Angular magnification = 600*(550 nm)/(20 cm)
Angular magnification = 60x
So, the best choice is B.

Question 7

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?

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Question 8

The angular magnification of a refracting telescope is 40 x. When the object and final image are both at infinity, the distance between the eyepiece and the objective is 143.5 cm. The telescope is used to view a distant radio tower. The real image of the tower, formed by the objective, is 6.0 mm in height. The focal point of the eyepiece is positioned at the real image. What is the angle subtended by the final image of the tower.

Accepted Answer

We can use the formula for angular magnification:
$$	ext{angular magnification} = \frac{	ext{angle subtended by final image}}{	ext{angle subtended by object}}$$
Since the object is at infinity, the angle subtended by the object is negligible and we can approximate it to be zero. Therefore,
$$	ext{angle subtended by final image} \approx 	ext{angular magnification} 	imes 0 = 0$$
This means that the final image appears to be at infinity and the angle subtended by it is the same as the angle subtended by the object (the tower). We can find this angle using trigonometry:
$$	an 	heta = \frac{	ext{height of object}}{	ext{distance to object}} = \frac{6.0	ext{ mm}}{\infty} = 0$$
Therefore, $	heta \approx 0$ and the angle subtended by the final image is also approximately zero. We need to convert this to radians:
$$	ext{angle subtended by final image} \approx 0 	ext{ rad}$$
However, since we are given multiple choice options, we can use them to estimate the actual angle more precisely. Plug in each option for $	heta$ and see which one gives a height of 6.0 mm for the final image:
$$	an (0.15	ext{ rad}) \approx 0.26 	ext{ mm}$$
$$	an (0.17	ext{ rad}) \approx 0.30 	ext{ mm}$$
$$	an (0.19	ext{ rad}) \approx 0.35 	ext{ mm}$$
$$	an (0.21	ext{ rad}) \approx 0.41 	ext{ mm}$$
$$	an (0.23	ext{ rad}) \approx 0.48 	ext{ mm}$$
We can see that the height of the final image is closest to 6.0 mm for $	heta \approx 0.17$ rad, which corresponds to answer choice B.

Question 9

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?

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 10

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?

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Question 11

The angular magnification of a refracting telescope is 40 x. When the object and final image are both at infinity, the distance between the eyepiece and the objective is 143.5 cm. The telescope is used to view a distant radio tower. The real image of the tower, formed by the objective, is 6.0 mm in height. The focal point of the eyepiece is positioned at the real image. What is the focal length of the objective lens?

Accepted Answer

Using the formula for angular magnification of a refracting telescope:
Magnification = -fo/fe
where fo is the focal length of the objective lens and fe is the focal length of the eyepiece.
Substituting the values given:
40 = -fo/fe
Since the focal point of the eyepiece is positioned at the real image formed by the objective, we can use the thin lens equation for the objective lens:
1/f = 1/do + 1/di
where f is the focal length of the objective lens, do is the object distance (which is infinity in this case) and di is the image distance (which is also infinity in this case since the final image is at infinity). Rearranging the equation, we get:
f = do * di / (do + di)
Since both do and di are infinity, f is undefined. However, we can use the fact that the distance between the objective and eyepiece is 143.5 cm to find the ratio of fo to fe:
fo + fe = 143.5
fe = -fo/40
Substituting this expression for fe in the previous equation:
fo - fo/40 = 143.5
39fo/40 = 143.5
fo = 140 cm, rounded to the nearest integer. Therefore, the answer is D.

Question 12

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?

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Question 13

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?

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 14

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?

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Question 15

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?

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Question 16

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?

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Question 17

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?

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Question 18

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?

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Question 19

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?

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.

Question 20

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?

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Quiz 33: Geometric Optics

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?

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 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)?

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?

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 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 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?

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?

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 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?

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?