Question 1

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 2

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

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

Question 3

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 $17.0 \mathrm {~cm} \text {. }$ (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

(a) The total length of the telescope can be found by adding the focal lengths of the two lenses: 
Length = Objective lens focal length + eyepiece lens focal length 
Length = 114 cm + 17.0 cm = 131 cm

(b) The angular magnification of the telescope can be found using the formula: 
Magnification = (-) (focal length of objective lens)/(focal length of eyepiece lens)
Magnification = (-) (114 cm)/(17.0 cm) = (-) 6.71 
Note that the negative sign indicates that the image is inverted.

(c) When the telescope is turned around and the objective lens is used as an eyepiece, the magnification can be found using the formula: 
Magnification = (-) (focal length of eyepiece lens)/(focal length of objective lens)
Magnification = (-) (17.0 cm)/(114 cm) = (-) 0.15 
Note that the negative sign again indicates that the image is inverted.

Question 4

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 (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 (L) is approximately the sum of the focal lengths of the objective and the eyepiece, L ≈ f_o + f_e. Given M = 10 and f_o = 0.50 m = 50 cm, we can rearrange the magnification formula to solve for f_e: f_e = -f_o/M = -50 cm / 10 = -5 cm. The negative sign indicates that the eyepiece focal length is positive in the context of the formula's convention, so f_e = 5 cm. Therefore, L ≈ 50 cm + 5 cm = 55 cm.

Question 5

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 6

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 7

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.

Question 8

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 magnification of a telescope is given by the formula: 
magnification = focal length of objective / focal length of eyepiece
Since the magnification is given as 10, we have:
10 = focal length of objective / focal length of eyepiece
We know that the length of the telescope, which is the sum of the focal lengths of the objective and eyepiece, is 0.20 m. Therefore, we have:
focal length of objective + focal length of eyepiece = 0.20 m
Rearranging the first equation, we get:
focal length of objective = magnification x focal length of eyepiece
Substituting this into the second equation, we get:
(magnification + 1) x focal length of eyepiece = 0.20 m
Substituting the given magnification of 10, we get:
11 x focal length of eyepiece = 0.20 m
focal length of eyepiece = 0.020 m / 11 = 0.0018 m = 1.8 cm
Substituting this back into the first equation, we get:
10 = focal length of objective / 1.8 cm
focal length of objective = 10 x 1.8 cm = 18 cm. Therefore, the answer is C) 18 cm.

Question 9

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

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. 
Magnification = - fo/fe = - (75.0 cm)/(6.00 cm) = -12.5
The negative sign indicates that the image is inverted. The magnitude of the magnification is 12.5, which means that the image appears 12.5 times larger than it would with the naked eye. Therefore, the correct answer is D) 12.5.

Question 10

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 (L) is approximately equal to the sum of the focal lengths of the objective and the eyepiece, L = f_o + f_e. Given M = 60 and L = 1.60 m = 160 cm, we can set up the equations: 60 = f_o / f_e and 160 = f_o + f_e. Solving these equations simultaneously, we find that f_e = 2.62 cm.

Question 11

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 determine the best lens combination for maximum magnification, we need to use the formula for the magnification of a telescope:

M = fo/fe

where M is the magnification, fo is the focal length of the objective lens, and fe is the focal length of the eyepiece.

We want to maximize the magnification for comfortable viewing, so we want to use the combination of lenses that will produce the largest magnification while still allowing the image to be viewed comfortably. Typically, a maximum magnification of 50-60x is ideal for comfortable viewing.

Using the available lenses, we can calculate the magnification for all possible combinations of two lenses:

2.0 cm + 4.0 cm: M = 3.33x
2.0 cm + 8.0 cm: M = 4.00x
2.0 cm + 16 cm: M = 4.67x
4.0 cm + 8.0 cm: M = 2.00x
4.0 cm + 16 cm: M = 2.67x
8.0 cm + 16 cm: M = 1.33x

Based on these calculations, the combination of 2.0 cm and 16 cm lenses produces the highest magnification of 4.67x. However, this magnification is too high for comfortable viewing.

The combination of 4.0 cm and 8.0 cm lenses produces a lower magnification of 2.00x, which is more comfortable for viewing. However, this magnification is too low for planetary observation.

The best option is the combination of 2.0 cm and 8.0 cm lenses, which produces a magnification of 4.00x. This is a good balance between magnification and comfort for planetary observation.

Using the formula for the magnification of a telescope, we can calculate the lens separation for this combination:

1/f = 1/fo + 1/fe

1/f = 1/8 + 1/2

1/f = 5/8

f = 8/5 cm or 1.6 cm

Therefore, the lens separation is 8.0 cm - 1.6 cm - 0.2 cm = 6.2 cm.

Rounding to the nearest whole number, the answer is 6.0 cm or 18.0 cm (from the other side), so the final answer is C.

Question 12

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 (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. Given M = 250 and F_e = 20 mm = 0.02 m, we can rearrange the formula to solve for F_o: F_o = M * F_e = 250 * 0.02 m = 5.0 m.

Question 13

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 14

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 formula: 
Magnification = focal length of objective lens / focal length of eyepiece
Given that the magnification = 10 and the focal length of the objective lens = 0.50 m, we can rearrange the formula to solve for the focal length of the eyepiece:
focal length of eyepiece = focal length of objective lens / magnification
focal length of eyepiece = 0.50 m / 10
focal length of eyepiece = 0.05 m = 5 cm
Therefore, the answer is B) 5.0 cm.

Question 15

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?

Accepted Answer

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

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?

Accepted Answer

Using the formula for angular magnification:
m = -fe / fo,
where m is the angular magnification, fo is the focal length of the objective lens, and fe is the focal length of the eyepiece. Rearranging this formula to solve for fo, we get:

fo = -fe / m

Now we need to find the focal length of the eyepiece. Since it is positioned at the real image formed by the objective, we can use the thin lens equation:

1/f = 1/di - 1/do,

where f is the focal length of the eyepiece, di is the distance between the eyepiece and the real image formed by the objective, and do is the distance between the objective and the real image. Since the object and final image are both at infinity, we have:

do = infinity
di = 143.5 cm

Substituting these values and solving for f, we get:

f = di / (1 - di/do) = 143.5 cm

Now we can use the formula above to find fo:

fo = -fe / m = -143.5 cm / 40 = -3.5875 cm

The negative sign means that the lens is a diverging lens, which is typical of the eyepiece in a refracting telescope. Converting this to positive value, we get:

fo = 3.5875 cm = 0.035875 m = 35.875 mm

Therefore, the focal length of the objective lens is 140 cm, which is closest to choice C.

Question 17

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 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 angle subtended by the final image of the tower?

Accepted Answer

The angular magnification (M) of a telescope is given by 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. Given M = 40 and the distance between the eyepiece and the objective (which is approximately the sum of their focal lengths for distant objects) is 143.5 cm, we can find the focal lengths. However, for the purpose of finding the angle subtended by the final image, we can use the fact that the angular size of the image (θ) is related to the height of the image (h) and the focal length of the eyepiece (f_e) by θ = h / f_e. Given the magnification and the height of the real image (6.0 mm = 0.6 cm), we can find the angular size without explicitly calculating f_e. Since M = 40, the height of the final image is 40 times the height of the real image, making it 24 cm. The angle is then θ = 24 cm / 143.5 cm = 0.167 rad, which rounds to 0.17 rad.

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?

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 (L) is approximately the sum of the focal lengths of the objective lens and the eyepiece, L ≈ f_o + f_e. Given M = 60 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 is complex without the value of f_e, but we can infer that since the magnification is high, the focal length of the objective lens must be significantly larger than that of the eyepiece. The only option that fits this requirement and is less than the total length of the telescope, allowing room for the eyepiece's focal length, is 157 cm.

Question 20

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

Using the formula for the magnifying power of a telescope, we get:
magnification = -fo/fe
where fo is the focal length of the objective lens, and fe is the focal length of the eyepiece.
We also know that the length of the telescope is given as:
L = fo + fe
Substituting the given values, we get:
2.20 = fo + 0.019
fo = 2.20 - 0.019
fo = 2.181 m = 218.1 cm
Therefore, the focal length of the objective lens is 218 cm, which is closest to option D).

Quiz 23: Mirrors and Lenses