Question 1

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 2

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?

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 3

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

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 4

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?

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 5

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?

Accepted Answer

The total length of the telescope is the sum of the focal lengths of the objective and eyepiece lenses: $114 \, \text{cm} + 17 \, \text{cm} = 131 \, \text{cm}$. The angular magnification is given by the ratio of the focal lengths: $ \frac{114}{17} \approx 6.71$. When reversed, the magnification is the reciprocal: $ \frac{17}{114} \approx 0.15$.

Question 6

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?

Accepted Answer

Using the formula for magnification, M = - fo / fe,
where fo = focal length of objective lens and fe = focal length of eyepiece.

Substituting the given values, we get:
M = - 75.0 cm / 6.00 cm = -12.5

Since magnification is a ratio, it does not have a unit. The negative sign indicates that the image is inverted. Therefore, the magnification of the telescope is 12.5 (or 12.5x).

Question 7

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?

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 8

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?

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 9

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?

Accepted Answer

The formula for magnification is M = fo/fe, where M is the magnification, fo is the focal length of the objective, and fe is the focal length of the eyepiece. Since the magnification is given as 10, we can assume that the focal length of the eyepiece is negligible (fe ≈ 0). Therefore, M ≈ fo/0, which means that fo ≈ M × 0 = 10 × 0.20 m = 2.0 m or 20 cm. However, we need to take into account the length limitation of the telescope, which is 0.20 m. To get a usable telescope length, we can use the formula L = fo + fe, where L is the telescope length. Since fe ≈ 0, we can assume that L ≈ fo, which means that fo ≤ 0.20 m. Therefore, the approximate focal length of the objective is fo ≈ 0.18 m or 18 cm, which is closest to choice C.

Question 10

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?

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Question 11

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"?

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Question 12

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?

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 13

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?

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 14

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?

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 15

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?

Accepted Answer

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Question 16

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?

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. In this case, the magnification is 10, and the focal length of the objective lens is 0.50 m. Therefore, the focal length of the eyepiece can be found by dividing the focal length of the objective lens by the magnification: 0.50 m / 10 = 0.05 m. The total length of the telescope is equal to the sum of the focal lengths of the objective lens and the eyepiece, plus the distance between them (which is usually about the sum of their diameters). Assuming the eyepiece has a diameter of about 2 cm, the total length of the telescope is approximately 0.50 m + 0.05 m + 0.02 m = 0.57 m, or 57 cm. However, since the answer choices are all in centimeters, the closest choice is B, which is 55 cm.

Question 17

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?

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Question 18

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?

Accepted Answer

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

Question 19

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?

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 20

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?

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.

Quiz 19: Optical Instruments