Question 1

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 formula for angular magnification is M = Fo/Fe, where Fo is the focal length of the objective lens and Fe is the focal length of the eyepiece. Rearranging the formula, we get Fo = M x Fe. 
Substituting the given values, we get Fo = 250 x 0.02 m = 5 m. Therefore, the answer is D) 5.0 m.

Question 2

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

The magnification of a telescope is given by the formula:
Magnification = - (focal length of objective lens) / (focal length of eyepiece)
Plugging the given values in the formula, we get:
Magnification = - (75.0 cm) / (6.00 cm) = -12.5
The negative sign indicates that the image formed by the eyepiece is inverted. Therefore, the magnification of the telescope is 12.5 times.

Question 3

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 4

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

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

Question 5

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

Using the formula for a telescope: 
magnification = focal length of objective lens / focal length of eyepiece
We can rearrange the formula to solve for the focal length of the objective lens:
focal length of objective lens = magnification x focal length of eyepiece

The magnification of a telescope is given by:
magnification = - (length of telescope / focal length of eyepiece)

Substituting the given values:
magnification = - (2.20 m / 1.9 cm)

Note that we need to convert centimeters to meters to get the units to match:
magnification = - (2.20 m / 0.019 m) = - 115.8

Now we can plug in the values for magnification and focal length of eyepiece:
focal length of objective lens = (-115.8) x (0.019 m) = -2.2042 m

Since the focal length must be positive, we take the absolute value to get:
focal length of objective lens = 2.2042 m ≈ 218 cm

Therefore, the best choice is D (218 cm).

Question 6

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 formula for magnification is M = -f1/f2, where f1 is the focal length of the objective and f2 is the focal length of the eyepiece. If we assume a standard eyepiece focal length of 0.02 m, we can solve for f1:

10 = -f1/0.02
f1 = -0.2 m/10 = -0.02 m

Since f1 must be positive, the focal length of the objective is approximately 0.18 m or 18 cm. Therefore, the best choice is C.

Question 7

An astronomical refracting telescope is made from two lenses. The objective lens has a focal length of $114 \mathrm {~cm} \text {, }$ and the eyepiece lens has a focal length of $17.0 \mathrm {~cm} .$ (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 
$\quad$(b) 6.71
$\quad$(c) 0.15
B) (a) 97 cm 
$\quad$(b) 0.15
$\quad$(c) 6.71
C) (a) 262 cm 
$\quad$(b) 0.87
$\quad$(c) 1.15
D) (a) 194 cm
$\quad$(b) 1.18
$\quad$(c) 0.85

Accepted Answer

(a) The total length of the telescope is the sum of the focal lengths of the lenses. 
$L = f_\text{o} + f_\text{e} = 114\text{ cm} + 17.0\text{ cm} = 131\text{ cm}$ 
(b) The angular magnification of the telescope is given by $M = - \frac{f_\text{o}}{f_\text{e}}$. 
$M = - \frac{114\text{ cm}}{17.0\text{ cm}} = -6.71$ (Note that the negative sign indicates that the image is inverted.) 
(c) The magnification by turning the telescope around and looking through the objective lens first is given by $M' = \frac{f_\text{o}}{f_\text{e}}$. 
$M' = \frac{114\text{ cm}}{17.0\text{ cm}} = 6.71$

Question 8

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 9

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 10

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.

Question 11

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 12

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 13

The objective lens of a telescope has a focal length of 1.90 m. When viewed through this telescope, the Moon appears 5.25 times larger than normal. How far apart are the objective lens and the eyepiece when this instrument is focused on the Moon?

Accepted Answer

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

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

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 formula for the magnification of a telescope is M = f' / f, where M is the magnification, f' is the focal length of the eyepiece, and f is the focal length of the objective lens. Since the magnification is given as 10 and the focal length of the objective lens is 0.50 m, we can solve for the focal length of the eyepiece:
M = f' / f
10 = f' / 0.50
f' = 5.00 m
The total length of the telescope is the sum of the focal lengths of the objective lens and the eyepiece:
L = f + f'
L = 0.50 + 5.00
L = 5.50 m
However, we usually want to express the length of the telescope in centimeters, not meters. Converting from meters to centimeters, we get:
L = 550 cm
Rounding to the nearest 5 cm, we get:
L ≈ 55 cm
Therefore, the best choice is B.

Question 16

The angular magnification of a refracting telescope is 40. 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) 138 cm
B) 139 cm
C) 140 cm
D) 141 cm

Accepted Answer

The answer of The angular magnification of a refracting telescope...

Question 17

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 18

Mary builds a simple refracting telescope using two lenses of focal length 80 cm and 5.0 cm. The respective diameters are 60 mm and 10 mm. How many degrees wide does the moon appear through the telescope if it appears only 1/2 degree wide with the "naked eye"?

Accepted Answer

The answer of Mary builds a simple refracting telescope using...

Question 19

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 20

A student wishing to build a telescope has available an eyepiece of focal length -5.0 cm. What focal length objective is needed to obtain a magnification of 10?

Accepted Answer

The answer of A student wishing to build a telescope...

Quiz 19: Optical Instruments

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 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 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?

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

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?

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 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?

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 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 objective lens of a telescope has a focal length of 1.90 m. When viewed through this telescope, the Moon appears 5.25 times larger than normal. How far apart are the objective lens and the eyepiece when this instrument is focused on the Moon?

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 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?

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?

Mary builds a simple refracting telescope using two lenses of focal length 80 cm and 5.0 cm. The respective diameters are 60 mm and 10 mm. How many degrees wide does the moon appear through the telescope if it appears only 1/2 degree wide with the "naked eye"?

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 student wishing to build a telescope has available an eyepiece of focal length -5.0 cm. What focal length objective is needed to obtain a magnification of 10?