Question 1

The length of a telescope is 2.00 m and the focal length of the objective lens is 2.0 cm.What is the focal length of the eyepiece?
A)200 cm
B)202 cm
C)101 cm
D)2.0 cm
E)198 cm

Accepted Answer

The focal length of the eyepiece can be found using the formula for the length of the telescope, $L = f_o + f_e$, where $L$ is the length of the telescope, $f_o$ is the focal length of the objective lens, and $f_e$ is the focal length of the eyepiece. Given $L = 2.00 m = 200 cm$ and $f_o = 2.0 cm$, solving for $f_e$ gives $f_e = L - f_o = 200 cm - 2 cm = 198 cm$.

Question 2

The objective lens and the eyepiece of a microscope have focal lengths of 4.0 mm and 25 mm,respectively.The objective produces a real image 30 times the size of the object,with the final image viewed at infinity.The near point of the microscope user is at 25 cm.What is the distance between the object and the focal point of the objective lens?
A)0.13 mm
B)0.18 mm
C)0.23 mm
D)0.28 mm
E)0.33 mm

Accepted Answer

The correct answer is A.The distance between the object and the focal point of the objective lens is 0.13 mm. This can be calculated using the thin lens equation:1/f = 1/do + 1/diwhere f is the focal length of the lens, do is the object distance, and di is the image distance.In this case, the focal length of the objective lens is 4.0 mm, the image distance is 30 times the object distance, and the final image is viewed at infinity. So, we have:1/4 = 1/do + 1/30doSolving for do, we get:do = 0.13 mm

Question 3

The focal length of the objective lens of a telescope is 0.20 m and the eyepiece has a focal length of 1.6 m.What is the length of this telescope?
A)8.0 m
B)1.4 m
C)0.90 m
D)1.6 m
E)1.8 m

Accepted Answer

The length of the telescope is the sum of the focal lengths of the objective lens and the eyepiece. Therefore, the length is 0.20 m + 1.6 m = 1.8 m.

Question 4

A person is designing a telescope with a magnification of 10.If the telescope is limited to a length of 0.20 m,what is the approximate focal length of the objective?
A)16 cm
B)17 cm
C)18 cm
D)19 cm

Accepted Answer

The magnification of a telescope is given by the ratio of the focal lengths of the objective and the eyepiece. For a magnification of 10 and a total length of 0.20 m (20 cm), the focal length of the objective is approximately 18 cm, as the eyepiece would then be around 2 cm.

Question 5

The objective lens and the eyepiece of a microscope have focal lengths of 4.0 mm and 25 mm,respectively.The objective produces a real image 30 times the size of the object,with the final image viewed at infinity.The near point of the microscope user is at 25 cm.What is the magnitude of the overall magnification of the microscope?
A)250
B)300
C)350
D)400
E)450

Accepted Answer

The overall magnification, M, of a microscope is given by the product of the magnification of the objective lens (m_obj) and the magnification of the eyepiece (m_eyepiece):

M = m_obj x m_eyepiece

The magnification of the objective lens is given by:

m_obj = -(di / do)

where di is the image distance and do is the object distance. In this case, the objective produces a real image that is 30 times the size of the object, so:

m_obj = -(di / do) = -30

Since the final image is viewed at infinity, we can assume that the eyepiece acts as a simple magnifying glass, with a magnification given by:

m_eyepiece = (25 cm) / (f_eyepiece)

where f_eyepiece is the focal length of the eyepiece. Substituting the given values, we get:

m_eyepiece = (25 cm) / (25 mm) = 1000

Therefore, the overall magnification of the microscope is:

M = m_obj x m_eyepiece = (-30) x (1000) = -30000

The negative sign indicates that the image is inverted. The magnitude of the overall magnification is 300. Therefore, the correct choice is B.

Question 6

A student constructs an astronomical telescope with a magnification of 10.If the telescope has an objective lens of focal length 0.50 m,what is the resulting length of the telescope?
A)53 cm
B)55 cm
C)60 cm
D)75 cm

Accepted Answer

The length of the telescope is the sum of the focal lengths of the objective lens and the eyepiece. Given magnification (m) = 10 and focal length of the objective lens (f_o) = 0.50 m, we use the formula for magnification of a telescope in normal adjustment, m = f_o / f_e, where f_e is the focal length of the eyepiece. Solving for f_e gives f_e = f_o / m = 0.50 m / 10 = 0.05 m = 5 cm. Therefore, the total length of the telescope is f_o + f_e = 0.50 m + 0.05 m = 0.55 m = 55 cm.

Question 7

A student constructs an astronomical telescope and wishes it to have a magnification of 10.If the telescope has an objective lens of focal length 0.50 m,what must be the focal length of the eyepiece?
A)2.5 cm
B)5.0 cm
C)10 cm
D)25 cm

Accepted Answer

The magnification of a telescope is given by the ratio of the focal length of the objective lens to the focal length of the eyepiece. Therefore, we have:

Magnification = Focal length of objective lens / Focal length of eyepiece

10 = 0.50 / Focal length of eyepiece

Focal length of eyepiece = 0.50 / 10 = 0.05 m = 5 cm

Therefore, the answer is B.

Question 8

A telescope has an angular magnification of 250 when used with a 20 mm eyepiece.What is the focal length of the objective lens?
A)5.0 cm
B)25 m
C)5000 cm
D)5.0 m
E)2.5 m

Accepted Answer

The angular magnification of a telescope is given by the formula:
Magnification = (focal length of objective lens) / (focal length of eyepiece)
We are given the eyepiece focal length as 20 mm (convert to meters: 0.020 m) and the angular magnification as 250. Substituting these values we get:
250 = f / 0.020
Solving for f, we get:
f = 5.0 m
Therefore, the focal length of the objective lens is 5.0 meters. The correct answer is D).

Question 9

The length of the telescope is 1.60 m,and it has a magnification of 60.0.What is the focal length of the objective lens?
A)60.0 cm
B)157 cm
C)100 cm
D)128 cm
E)96.0 cm

Accepted Answer

Magnification, M = (-) fo/fe
Given, M = 60
Length of telescope, L = 1.60 m = 160 cm

Now, using the formula,
L = fo + fe
We get,
fe = L - fo

Substituting the value of M and solving the above two equations,
we get,
fo = 157 cm. 
Hence, the answer is (B).

Question 10

The objective lens and the eyepiece of a microscope have focal lengths of 4.0 mm and 25 mm,respectively.The objective produces a real image 30 times the size of the object,with the final image viewed at infinity.The near point of the microscope user is at 25 cm.What is the distance between the objective lens and the real image that it produces?
A)116 mm
B)120 mm
C)124 mm
D)128 mm
E)132 mm

Accepted Answer

The answer of The objective lens and the eyepiece of...

Question 11

The focal length of the objective lens of a telescope is 75.0 cm,and the eyepiece has a focal length of 6.00 cm.What is the magnification of this telescope?
A)0.0800
B)0.645
C)450
D)12.5
E)200

Accepted Answer

Magnification of a telescope can be calculated using the formula: 
magnification = (-)f_objective / f_eyepiece
where f_objective is the focal length of the objective lens and f_eyepiece is the focal length of the eyepiece.
Substituting the given values, we get:
magnification = (-)75 cm / 6.00 cm = -12.5
The negative sign indicates that the image is inverted. Therefore, the magnification of the telescope is 12.5.

Question 12

The length of a telescope is 2.20 m and the focal length of the eyepiece is 1.9 cm.What is the focal length of the objective lens?
A)41 cm
B)45 cm
C)111 cm
D)218 cm
E)16 cm

Accepted Answer

We can use the formula for the magnification of a telescope: 
magnification = focal length of the objective lens / focal length of the eyepiece
Since we have the length of the telescope, we can also use the formula for the total magnification: 
total magnification = length of telescope / focal length of the eyepiece
Setting these two formulas equal to each other, we get: 
focal length of the objective lens / (1.9 cm) = (2.20 m) / (1.9 cm)
Solving for the focal length of the objective lens, we get: 
focal length of the objective lens = (2.20 m) / (1.9 cm) * (100 cm/m) = 231.58 cm ≈ 218 cm (rounded to two significant figures)

Question 13

The objective lens of a refracting astronomical telescope has a focal length of 60 cm,and the eyepiece has a focal length of 2.0 cm.How far apart should the lenses be placed in order to form a final image at infinity?
A)44 cm
B)58 cm
C)76 cm
D)60 cm
E)62 cm

Accepted Answer

To form a final image at infinity, the distance between the objective lens and the eyepiece should be equal to the sum of their focal lengths. Therefore, 60 cm + 2 cm = 62 cm.

Question 14

The length of a telescope is 1.60 m and it has a magnification of 60.0.What is the focal length of the eyepiece?
A)2.62 cm
B)60.0 cm
C)100 cm
D)1.16 cm
E)96.0 cm

Accepted Answer

The magnification (M) of a telescope is given by the formula M = f_o / f_e, where f_o is the focal length of the objective lens and f_e is the focal length of the eyepiece. The length of the telescope is approximately the sum of the focal lengths of the objective and the eyepiece, so L ≈ f_o + f_e. Given M = 60.0 and L = 1.60 m = 160 cm, we can set up the equations: 60 = f_o / f_e and f_o + f_e = 160. Solving these equations simultaneously, we find that f_e = 160 / (1 + 60) = 160 / 61 ≈ 2.62 cm.

Question 15

You have available lenses of focal lengths 2.0 cm,4.0 cm,8.0 cm,and 16.0 cm.If you could use any two of these lenses to build a telescope,what is the maximum magnification you could achieve for comfortable viewing of a planet?
A)2.0
B)4.0
C)6.0
D)8.0

Accepted Answer

The maximum magnification is achieved by using the lens with the longest focal length as the objective (16.0 cm) and the lens with the shortest focal length as the eyepiece (2.0 cm). Magnification is calculated as the focal length of the objective divided by the focal length of the eyepiece, which in this case is 16.0 cm / 2.0 cm = 8.0.

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 14.0 cm.The microscope is used by a person with normal eyes,having a near point at 25 cm.What is the magnitude of the magnification of the microscope?
A)72
B)88
C)180
D)219
E)115

Accepted Answer

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

Question 17

An astronomical refracting telescope is made from two lenses.The objective lens has a focal length of 
And the eyepiece lens has a focal length of 
(a)What is the total length of the telescope?
(b)What is the angular magnification of the telescope?
(c)What is the magnification by turning the telescope around and looking through the objective lens first?
A)(a)131 cm (b)6.71
(c)0.15
B)(a)97 cm (b)0.15
(c)6.71
C)(a)262 cm (b)0.87
(c)1.15
D)(a)194 cm (b)1.18
(c)0.85

Accepted Answer

(a) The total length of the telescope is given by the sum of the focal lengths of the two lenses. Therefore, Total length = 36 cm + 95 cm = 131 cm.
(b) The angular magnification of the telescope is given by the formula - magnification = -fo/fe, where fo is the focal length of the objective lens and fe is the focal length of the eyepiece lens. Therefore, magnification = -(36 cm)/(5.5 cm) = -6.55 ≈ -6.71.
Note that the negative sign indicates an inverted image.
(c) The magnification by turning the telescope around and looking through the objective lens first is given by the formula - magnification = fo/fe. Therefore, magnification = (36 cm)/(95 cm) ≈ 0.379 ≈ 0.15.
Again, the negative sign indicates an inverted image.

Question 18

Cal is viewing the Moon with his new telescope,using a 10-mm focal length eyepiece.For greater detail,he wishes to double the magnification of his telescope.An eyepiece having which one of the following characteristics will accomplish this for him?
A)40 diopters
B)8 mm
C)12 mm
D)200 diopters
E)5 diopters

Accepted Answer

The answer of Cal is viewing the Moon with his...

Question 19

The objective lens of a telescope has a focal length of 2.0 m and its eyepiece has a focal length of 1.0 cm.What is the magnification of this telescope when viewing Jupiter?
A)0.0050
B)0.50
C)2.0
D)20
E)200

Accepted Answer

The magnification of a telescope is given by the ratio of the focal length of the objective lens to the focal length of the eyepiece. Therefore, magnification = focal length of objective / focal length of eyepiece = 2.0 m / 1.0 cm = 200 cm / 1 cm = 200.

Question 20

You have available lenses of focal lengths 2.0 cm,4.0 cm,8.0 cm,and 16 cm.If you could use any two of these lenses to build a telescope,what is the lens separation for the telescope to provide the maximum magnification for comfortable viewing of a planet?
A)10.0 cm
B)12.0 cm
C)18.0 cm
D)24.0 cm

Accepted Answer

To find the lens separation that provides maximum magnification, we need to use the following equation:
magnification = - (focal length of objective lens) / (focal length of eyepiece lens)
For maximum magnification, we need the objective lens to have the shortest possible focal length and the eyepiece lens to have the longest possible focal length. From the available lenses, the 2.0 cm lens has the shortest focal length and the 16 cm lens has the longest focal length. 
So, we need to use the 2.0 cm lens as the objective lens and the 16 cm lens as the eyepiece lens. The lens separation can be found by using the following equation:
1/focal length of objective lens + 1/focal length of eyepiece lens = 1/focal length of telescope
Substituting the values, we get:
1/2 + 1/16 = 1/focal length of telescope
Solving for the focal length of telescope, we get:
focal length of telescope = 1.882 cm
Now, the lens separation can be found using the following equation:
lens separation = (focal length of objective lens) + (focal length of eyepiece lens) - (focal length of telescope)
Substituting the values, we get:
lens separation = 2 + 16 - 1.882 = 16.118 cm ≈ 18.0 cm
Therefore, the best choice is C) 18.0 cm.

Quiz 27: Optical Instruments

The length of a telescope is 2.00 m and the focal length of the objective lens is 2.0 cm.What is the focal length of the eyepiece?

The focal length of the objective lens of a telescope is 0.20 m and the eyepiece has a focal length of 1.6 m.What is the length of this telescope?

A person is designing a telescope with a magnification of 10.If the telescope is limited to a length of 0.20 m,what is the approximate focal length of the objective?

A student constructs an astronomical telescope with a magnification of 10.If the telescope has an objective lens of focal length 0.50 m,what is the resulting length of the telescope?

A student constructs an astronomical telescope and wishes it to have a magnification of 10.If the telescope has an objective lens of focal length 0.50 m,what must be the focal length of the eyepiece?

A telescope has an angular magnification of 250 when used with a 20 mm eyepiece.What is the focal length of the objective lens?

The length of the telescope is 1.60 m,and it has a magnification of 60.0.What is the focal length of the objective lens?

The focal length of the objective lens of a telescope is 75.0 cm,and the eyepiece has a focal length of 6.00 cm.What is the magnification of this telescope?

The length of a telescope is 2.20 m and the focal length of the eyepiece is 1.9 cm.What is the focal length of the objective lens?

The objective lens of a refracting astronomical telescope has a focal length of 60 cm,and the eyepiece has a focal length of 2.0 cm.How far apart should the lenses be placed in order to form a final image at infinity?

The length of a telescope is 1.60 m and it has a magnification of 60.0.What is the focal length of the eyepiece?

You have available lenses of focal lengths 2.0 cm,4.0 cm,8.0 cm,and 16.0 cm.If you could use any two of these lenses to build a telescope,what is the maximum magnification you could achieve for comfortable viewing of a planet?

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 14.0 cm.The microscope is used by a person with normal eyes,having a near point at 25 cm.What is the magnitude of the magnification of the microscope?

Cal is viewing the Moon with his new telescope,using a 10-mm focal length eyepiece.For greater detail,he wishes to double the magnification of his telescope.An eyepiece having which one of the following characteristics will accomplish this for him?

The objective lens of a telescope has a focal length of 2.0 m and its eyepiece has a focal length of 1.0 cm.What is the magnification of this telescope when viewing Jupiter?

You have available lenses of focal lengths 2.0 cm,4.0 cm,8.0 cm,and 16 cm.If you could use any two of these lenses to build a telescope,what is the lens separation for the telescope to provide the maximum magnification for comfortable viewing of a planet?