Question 1

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 angular magnification of a magnifying glass is given by $M=\frac{	heta'}{	heta}$, where $	heta'$ is the angle subtended by the image at the eye and $	heta$ is the angle subtended by the object at the eye. To find $	heta'$, we can use the thin lens formula $\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}$, where $f$ is the focal length, $d_o$ is the object distance, and $d_i$ is the image distance. Since the image is at the far point, $d_i=\infty$. Solving for $d_o$, we get $d_o=\frac{f}{2}$. Using the small angle approximation $	heta\approx	an	heta=\frac{h}{d_o}$, where $h$ is the height of the object, and the fact that the eye is relaxed, we can assume that the object is at the near point, so $d_o=25$ cm. Thus, $	heta=\frac{h}{25}$.

To find $	heta'$, we need to use the fact that the eye is relaxed, so the image is at the far point. Thus, $d_i=\infty$, and the magnifying glass must produce a virtual image at the near point. The magnification of a magnifying glass is given by $m=\frac{25}{f}$, where $f$ is the focal length. Since the image is virtual, the magnification is negative, so $m=-\frac{25}{4}=-6.25$. The angular magnification is then $M=\frac{	heta'}{	heta}=m+1=-5.25$. The negative sign indicates that the image is inverted.

The closest choice to this value is E, which gives an angular magnification of 6.3x. This is the best choice.

Question 2

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

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

Question 3

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

First we need to find the magnification of the telescope without the eyepiece:
$$M_{telescope}=-\frac{f_{obj}}{f_{eye}}=-\frac{180\ 	ext{cm}}{3\ 	ext{cm}}=-600$$
Since we have an eyepiece, we can find the total angular magnification by also including the magnification of the eyepiece:
$$M_{total}=M_{telescope}\cdot M_{eye}=-600\cdot \frac{|f_{obj}|}{f_{eye}}=-600\cdot \frac{180\ 	ext{cm}}{3\ 	ext{cm}}=-36000$$
However, we need to take the absolute value of the objective focal length because the angular magnification is always positive. Finally, we can convert this to a more common format:
$$M_{total}=\frac{	ext{angle subtended at eye with telescope}}{	ext{angle subtended at eye without telescope}}=-36000	imes$$
Therefore, the answer is B) 60x.

Question 4

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

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Question 5

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?
A) 137 cm
B) 138 cm
C) 139 cm
D) 140 cm
E) 141 cm

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 6

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

The answer of You have available lenses of focal lengths...

Question 7

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

Accepted Answer

The answer of The objective lens of a refracting telescope...

Question 8

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 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?
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 10

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)?

Accepted Answer

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Question 11

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

The answer of What is the focal length of a...

Question 12

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?

Accepted Answer

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

Question 13

The objective and the eyepiece of a refracting astronomical telescope have focal lengths of 320 cm and 4.0 cm, respectively. The telescope is used to view Neptune and the final image is set at infinity. The diameter of Neptune is 4.96 × 10⁷m and the distance from the earth at the time of observation is 4.4 × 10¹²m. What is the angle (in mrad) subtended by the final telescopic image of Neptune? A) 0.90 mrad B) 1.1 mrad C) 1.3 mrad D) 1.5 mrad E) 1.7 mrad

Accepted Answer

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Question 14

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

The answer of The distance between the object and the...

Question 15

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

Accepted Answer

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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?
A) 78 x
B) 94 x
C) 195 x
D) 234 x
E) 125 x

Accepted Answer

The answer of The eyepiece of a compound microscope has...

Question 17

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

Accepted Answer

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Question 18

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.
A) 0.15 rad
B) 0.17 rad
C) 0.19 rad
D) 0.21 rad
E) 0.23 rad

Accepted Answer

We can use the formula for angular magnification M = (-) fo/fe, where fo is the focal length of the objective and fe is the focal length of the eyepiece. Rearranging, we have fe = (-) fo/M.

Substituting in our values, we have fe = (-) fo/40.

We can also use the thin lens formula for the objective lens, 1/f = 1/do + 1/di, where f is the focal length, do is the object distance (which is infinity in this case), and di is the image distance. Since the final image is also at infinity, di = infinity and we can simplify the equation to f = do.

Substituting in our values, we have fo = 143.5 cm.

From the problem, we know that the eyepiece is positioned at the real image formed by the objective, so the final image is also at this location. This means the final image is the same height as the real image, 6.0 mm.

To find the angle subtended by the final image, we need to find the height of the final image at the eyepiece, h′, and then use the formula for angular size, θ = h′/fe.

Using the thin lens formula for the eyepiece, we have 1/fe = 1/h′ + 1/di, where di is the distance between the objective and the eyepiece, which we know is 143.5 cm. Since the final image is at infinity, di = infinity and we can simplify the equation to 1/fe = 1/h′.

Substituting in our values, we have fe = (-) fo/40 = (-) 143.5 cm/40 = (-) 3.59 cm. Solving for h′, we have 1/fe = 1/h′, so h′ = fe = (-) 3.59 cm.

Finally, substituting in our values for θ = h′/fe, we have θ = (-) 3.59 cm/(-) 3.59 cm = 1 rad = 57.3 degrees. Converting to radians, we get θ ≈ 0.17 rad, so the answer is B.

Question 19

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

Question 20

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

The total overall magnification of a compound microscope is given by:
Magnification = (magnification of objective lens) x (magnification of eyepiece)
The magnification of the objective lens is given by:
magnification of objective lens = focal length of objective lens / (distance between object and objective lens)
magnification of objective lens = 0.80 cm / (18.0 cm - 0.80 cm)
magnification of objective lens = 0.0465x
The magnification of the eyepiece is given by:
magnification of eyepiece = 1 + (near-point distance / focal length of eyepiece)
magnification of eyepiece = 1 + (25.0 cm / 2.3 cm)
magnification of eyepiece = 11.87x
Therefore, the total overall magnification is:
Magnification = (magnification of objective lens) x (magnification of eyepiece)
Magnification = (0.0465x) x (11.87x)
Magnification = 0.553x
The negative sign indicates that the image is inverted. Therefore, the best choice is C, which gives a total overall magnification of -200x.

Quiz 24: Optical Instruments

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?

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?

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 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)?

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 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?

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?