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Chapter 4: Problem 17

Generally CCDs have fewer picture elements (pixels) than do photographic plates, so if you want to image a large field, a single photograph might suffice, but you would need a number of CCD images. Suppose you had to make a \(3 \times 3\) square of \(\mathrm{CCD}\) images to cover your single photographic field, and that your photograph has a quantum efficiency of \(5 \%\) while your CCD has an efficiency of \(80 \%\). How long will the needed CCD images take relative to the time for the photograph?

Short Answer

Expert verified
The CCD images will take 1/144th of the time relative to the photograph.

Step by step solution

01

Identify relevant information

The problem states that a photograph requires a grid of CCD images (3x3), with the photograph's quantum efficiency being 5% and the CCD's efficiency being 80%.
02

Calculate the efficiency ratio

Determine the ratio of efficiencies between the photograph and CCD. This can be done by dividing the CCD's efficiency by the photograph's efficiency: \ \[ \text{Efficiency Ratio} = \frac{80\text{\textperthousand}}{5\text{\textperthousand}} = 16. \ \]
03

Calculate the total number of CCD images needed

Since the problem mentions you need a 3x3 grid of CCD images to cover the same area as a single photograph, the number of CCD images required is: \ \[ \text{Number of CCD images} = 3 \times 3 = 9. \ \]
04

Determine the total effective efficiency for CCD images

Given that there are 9 CCD images each with an efficiency ratio of 16, we calculate the total effective efficiency: \ \[ \text{Total Efficiency} = \text{Number of CCD Images} \times \text{Efficiency Ratio} = 9 \times 16 = 144. \ \] This indicates that the CCD images will be 144 times more efficient than the photograph.
05

Compute the relative time for CCD images compared to the photograph

To find the relative time taken for the CCD images compared to the photograph, we take the reciprocal of the total effective efficiency since higher efficiency results in less time needed: \ \[ \text{Relative Time} = \frac{1}{144}. \ \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Quantum Efficiency
Quantum efficiency is a critical concept in imaging technology. It refers to the ability of a device to convert incoming photons into usable electronic signals.
For instance, if a camera sensor has a quantum efficiency of 80%, it can convert 80% of the incoming light into an image.
Conversely, if it has a 5% efficiency, only 5% of that light gets utilized for image creation.
Higher quantum efficiency means better performance, particularly in low-light conditions.
This parameter plays a key role in comparing the effectiveness of CCDs and photographic plates.
CCD Imaging
CCD imaging refers to the use of Charge-Coupled Devices to capture images.
CCD sensors are widely used in digital cameras and astronomical telescopes.
These sensors are highly efficient due to their higher quantum efficiency compared to traditional photographic plates.
Despite requiring multiple images to cover a large field, CCDs offer significant advantages.
  • They provide higher resolution.
  • They have better sensitivity to light.
  • They allow for digital manipulation and enhancement of images.
Understanding CCD imaging is essential for grasping why they are preferred for many modern applications.
Photographic Plates
Photographic plates were commonly used before the advent of digital imaging.
They are large glass plates coated with light-sensitive emulsion.
While they cover a large field in a single shot, they have low quantum efficiency.
In the given exercise, the photographic plate has a 5% quantum efficiency, meaning it can only convert 5% of the light into a visible image.
Despite their lower efficiency, photographic plates were valuable for capturing wide-field images in astronomical applications.
However, their low quantum efficiency and the advent of digital technology have reduced their usage.
Efficiency Ratio
The efficiency ratio helps compare the effectiveness of two different imaging devices.
In the exercise, we see an efficiency ratio calculated as 16.
This comes from dividing the CCD's quantum efficiency (80%) by the photographic plate's quantum efficiency (5%).
An efficiency ratio of 16 means that a CCD is 16 times more efficient at converting light into an image compared to a photographic plate.
  • Higher efficiency ratios indicate better performance.
  • They also influence the relative imaging time required for capturing comparable images.
This concept is essential for evaluating which technology is more suited to specific imaging needs.
Relative Imaging Time
Relative imaging time is the time needed to capture an image with one device compared to another.
Higher efficiency means less time required to capture a high-quality image.
In our problem, the CCD's higher efficiency translates to a much shorter imaging time compared to the photographic plates.
Given an efficiency ratio of 144, the relative time for CCDs is calculated as \[ \text{Relative Time} = \frac{1}{144} = 0.00694... \].
This means that the CCD requires only about 0.7% of the time that a photographic plate would need to achieve the same imaging results.
This significant reduction in time is a primary reason for the preference of CCDs in many applications.

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Most popular questions from this chapter

(a) Using the fact that the limiting magnitude of the eye is \(6,\) derive an expression for the limiting magnitude for direct viewing with a telescope of diameter \(D\). (Ignore the effects of sky brightness.) (b) Use this result to derive an expression for the farthest distance at which a telescope of diameter \(D\) can be used to see an object of absolute magnitude \(M\).

Suppose that we use a reflector in the coudé focus, and each of the three mirrors reflects \(95 \%\) of the light. What fraction of the light is lost as a result of these three reflections?

The full Moon subtends an angle of approximately 30 arc min. How large would the image of the Moon be on your film if you used a \(500 \mathrm{mm}\) focal length lens for your camera?

Two infrared sources in the Orion Nebula are 500 pc from us and are separated by 0.1 pc. How large a telescope would you need to distinguish the sources at a wavelength of \(100 \mu \mathrm{m} ?\)

The sodium D lines in the Sun's spectrum are at wavelengths of 589.594 and \(588.997 \mathrm{nm}\) (a) If a grating has \(10^{4}\) lines/cm, how wide must the grating be to resolve the two lines in first order? (b) Under these conditions what is the angular separation between the two lines? (c) How would the results in (a) and (b) change for second order?
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