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Chapter 6: Problem 39

Consider a CCD with a quantum efficiency of 80 percent and a photographic plate with a quantum efficiency of 1 percent. If an exposure time of 1 hour is required to photograph a celestial object with a given telescope, how much observing time would be saved by substituting a CCD for the photographic plate?

Short Answer

Expert verified
Observing time saved by substituting a CCD is 59.25 minutes.

Step by step solution

01

- Understanding Quantum Efficiency

Quantum efficiency (QE) is the percentage of photons hitting the detector that are actually detected. For a CCD, QE is 80%, and for a photographic plate, it is 1%.
02

- Calculate the Efficiency Ratio

To find the efficiency ratio between the CCD and the photographic plate, divide the CCD's QE by the photographic plate's QE: \(\text{Efficiency Ratio} = \frac{80\text{%}}{1\text{%}} = 80 \)
03

- Determine Time Saved

Since the CCD is 80 times more efficient than the photographic plate, it requires 80 times less exposure time to capture the same amount of light. If the photographic plate requires 1 hour, then: \(\text{Time Saved} = 1 \text{ hour} - \frac{1 \text{ hour}}{80} = 1 \text{ hour} - 0.0125 \text{ hours} \)
04

- Convert Time Saved to Minutes

Convert the time saved from hours to minutes: \(\text{Time Saved in Minutes} = 1 \text{ hour} - 0.0125 \text{ hours} = 60 \text{ minutes} - 0.75 \text{ minutes} = 59.25 \text{ minutes} \)

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Key Concepts

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

CCD
A Charge-Coupled Device (CCD) is a highly sensitive detector used in various imaging technologies. It is popular in telescopes, cameras, and medical imaging devices.
CCDs have many advantages:
  • They offer high quantum efficiency, meaning they can detect a significant percentage of incoming photons.
  • They're capable of capturing very faint objects with great accuracy, making them ideal for astronomical observations.
  • Each pixel in a CCD chip is capable of detecting light and storing it as electrical charges proportional to the photon intensity.
These characteristics make CCDs a crucial component in modern telescopic observations.
photographic plate
Photographic plates were one of the earliest methods for capturing astronomical images. They consist of a glass plate coated with a light-sensitive emulsion.
Although they were revolutionary in their time, they have several limitations:
  • Low quantum efficiency, around 1%, meaning they detect very few incoming photons.
  • They require long exposure times to gather sufficient light, making them less efficient than modern digital counterparts.
  • Handling and processing photographic plates involve chemical development, which is cumbersome and time-consuming.
These limitations have led to the decline in their use, with CCDs becoming the preferred choice for astronomical imaging.
exposure time
Exposure time is the duration that a camera's sensor or photographic plate is exposed to light.
Longer exposure times allow detectors to capture more light, making faint objects more visible. However, longer exposure times also mean more time observing the same region of the sky, which can be impractical.
In the given exercise, the exposure time for a photographic plate is 1 hour. This can be drastically reduced to almost 0.75 minutes if using a CCD, due to its higher quantum efficiency.
light detection efficiency
Light detection efficiency refers to how effectively a detector can convert incoming photons into a measurable signal. The key term 'quantum efficiency' (QE) quantifies this efficiency.
Quantum efficiency is defined as the percentage of photons hitting the detector that result in detected signals:
  • High QE means a greater portion of incoming light is detected, leading to more accurate and sensitive measurements.
  • Low QE means most photons are lost, requiring longer exposure times to achieve the same signal quality.
  • In our exercise, the quantum efficiency of a CCD is 80%, while that of a photographic plate is just 1%.
Understanding this concept helps explain why CCDs save significant observing time compared to photographic plates.

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Most popular questions from this chapter

The angular resolution of a ground-based telescope is usually determined by a. diffraction. b. the focal length. c. refraction. d. atmospheric seeing.

One of the earliest astronomical CCDs had 160,000 pixels, each recording 8 bits ( 256 levels of brightness). A new generation of astronomical CCDs may contain a billion pixels, each recording 15 bits \((32,768\) levels of brightness). Compare the number of bits of data that each of these two CCD types produces in a single image.

A grating is able to spread white light out into a spectrum of colors because of the property of a. reflection. b. diffraction. c. dispersion. d. interference.

The two Keck 10 -meter telescopes, separated by a distance of 85 meters, can operate as an optical interferometer. What is its resolution when it observes in the infrared at a wavelength of 2 microns? a. 0.01 aresec b. 0.005 arcsec c. 0.2 arsec d. 0.05 arcsec

The diameter of the full Moon in the focal plane of an average amateur's telescope (focal length 1.5 meters) is \(13.8 \mathrm{mm}\) How big would the Moon be in the focal plane of a very large astronomical telescope (focal length 250 meters)?
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