College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Assistance with part 4
![You want to create a telescope with a resolving power of 0.100 arc seconds at a wavelength of 550 nm. What diameter (in m) do you need?
If you want to increase the light gathering power by a factor of 10, by what factor does the diameter need to increase?
What would the new resolving power be (in arc seconds)?
**Part 1 of 5**
The resolving power of a telescope is given by:
\[
\alpha = 2.06 \times 10^5 \frac{\lambda}{D}
\]
where both \(\lambda\) and \(D\) must have the same units.
---
**Part 2 of 5**
We need to solve this expression for the diameter. You should express your values in meters.
\[
D = 2.06 \times 10^5 \times \frac{0.00000055 \, \text{m}}{0.100 \, \text{arc seconds}}
\]
\[
D = 1.133 \approx 1.13 \, \text{m}
\]
---
**Part 3 of 5**
The light gathering power of a telescope depends on the diameter squared.
\[
\frac{LGP_{\text{new}}}{LGP_{\text{old}}} = \left(\frac{D_{\text{new}}}{D_{\text{old}}}\right)^2
\]
---
**Part 4 of 5**
If we want to increase the light gathering power by a factor of 10, we need the new telescope to have 10 times the light-gathering power of the original.
\[
LGP_{\text{new}} = 10 \times LGP_{\text{old}}
\]
Rearranging the equation for light-gathering power to solve for \(D_{\text{new}}\):
\[
D_{\text{new}} = \sqrt{\frac{LGP_{\text{new}}}{LGP_{\text{old}}}} \times D_{\text{old}}
\]
Plugging in for \(LGP_{\text{new}}\) gives:
\[
D_{\text{new}} = \sqrt{10} \times D_{\text{old}}
\]
And using the initial diameter, we can solve for the new diameter.
\[
D_{\text{new}} \approx 3.58 \, \text{m](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fda14446f-ef4d-47d5-9b14-b7a32b5fa1b1%2F74c71c32-4156-4b50-bda3-b1d134e225c8%2Fayzvtd5.jpeg&w=3840&q=75)
Transcribed Image Text:You want to create a telescope with a resolving power of 0.100 arc seconds at a wavelength of 550 nm. What diameter (in m) do you need?
If you want to increase the light gathering power by a factor of 10, by what factor does the diameter need to increase?
What would the new resolving power be (in arc seconds)?
**Part 1 of 5**
The resolving power of a telescope is given by:
\[
\alpha = 2.06 \times 10^5 \frac{\lambda}{D}
\]
where both \(\lambda\) and \(D\) must have the same units.
---
**Part 2 of 5**
We need to solve this expression for the diameter. You should express your values in meters.
\[
D = 2.06 \times 10^5 \times \frac{0.00000055 \, \text{m}}{0.100 \, \text{arc seconds}}
\]
\[
D = 1.133 \approx 1.13 \, \text{m}
\]
---
**Part 3 of 5**
The light gathering power of a telescope depends on the diameter squared.
\[
\frac{LGP_{\text{new}}}{LGP_{\text{old}}} = \left(\frac{D_{\text{new}}}{D_{\text{old}}}\right)^2
\]
---
**Part 4 of 5**
If we want to increase the light gathering power by a factor of 10, we need the new telescope to have 10 times the light-gathering power of the original.
\[
LGP_{\text{new}} = 10 \times LGP_{\text{old}}
\]
Rearranging the equation for light-gathering power to solve for \(D_{\text{new}}\):
\[
D_{\text{new}} = \sqrt{\frac{LGP_{\text{new}}}{LGP_{\text{old}}}} \times D_{\text{old}}
\]
Plugging in for \(LGP_{\text{new}}\) gives:
\[
D_{\text{new}} = \sqrt{10} \times D_{\text{old}}
\]
And using the initial diameter, we can solve for the new diameter.
\[
D_{\text{new}} \approx 3.58 \, \text{m
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