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StillLearningToronto

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**(Sorry for Length)**

1. Homework Statement

1. Homework Statement

You have decided to use a CCD camera to check if a 16th magnitude quasar is variable. You can presume that all of the light of your quasar falls on one pixel. You know that a star with a magnitude of 0 would deliver 1 × 10^9 photons/second to one pixel.

The quantum efficiency of the CCD is 90%. The dark current is 2 electrons/second/pixel, and the maximum signal that a pixel can hold (the full well) is 120,000 electrons. The bias level is 1000 ADUs, and the maximum signal that the A/D converter can read is 65,535 ADUs. If the gain of the CCD is 2 electrons per ADU, what is the maximum length of time (in minutes) that you should hold the shutter of the camera open?

a) the rate at which electrons are generated by the quasar flux

b) the maximum number of electrons recordable by the A/D converter

c) the maximum number of electrons that can be accumulated in a well before it responds non-linearly to flux and the corresponding number of electrons that must be read out taking into account the bias

d) which of the two maxima limits the exposure time

e) in symbolic terms, the total signal read out from a pixel after an exposure of length t taking into account the quasar flux F (per unit time), the dark current D (per unit time), and the bias B.

f) the limiting exposure time, after setting the total signal read out to be the identified maximum

## Homework Equations

m2-m1=-2.5log(F2/F1)

Electrons/second=F2*90%

A/D max=ADU*gain-bias

## The Attempt at a Solution

- a) the rate at which electrons are generated by the quasar flux

m2-m1=-2.5log(F2/F1)

16-0=-2.5log(F2/(1*109photons/second)

F2=3.98*102 photons/second

Electrons/second=F2*90%

e/s=3.98*102 photons/second

e/s=358.2

:.The total number of electrons per second is 358.2

First I took the magnitude of the comparison star, with the magnitude of my quasar (16th) to solve for the flux of my quasar. I then multiplied that by the quantum efficiency (90%) to get my total electrons per second hitting the pixel in question.

b) the maximum number of electrons recordable by the A/D converterA/D max=ADU*gain-bias

A/D max = 65535ADU*(1000ADU*2electrons)-2000electrons

A/D max= 129070 electrons

To find the A/D converter’s max number of electrons I took the ADU amount, multiplied it by the gain (2 electrons per ADU) and subtracted the bias of 2000 electrons.c) the maximum number of electrons that can be accumulated in a well before it responds non-linearly to flux (refer to Chapter 3), and the corresponding number of electrons that must be read out taking into account the bias

MAX well capacity for full well

MAX well capacity=120,000 electrons-bias

MAX well capacity=118,000 electrons

*At the recomended 70%

MAX well capacity=120,000 electrons*70%-bias

MAX well capacity=82,000 electrons

Here i did two options: to have a full well, or the recommended 70% to prevent non-linear. I took the max well capacity (120,000 electrons) multiplied that by the 70% to not “overfill” the well, and then subtracted the bias (2000 electrons.)

*Point to note, i was not sure if you subtracted the bias before or after multiplying by 70%.d) which of the two maxima limits the exposure time

The two maxima that limits the exposure time is the Max well limit and the max amount of ADU the A/D converter can read.

The max well limit because once a well fills up too much (past 70%) you can start to get too much noise/ non-linear responding which would corrupt proper data.

The gain because it sets the pace as to how much

The gain because it limits the amount of electrons per ADU. It is also an adjustable parameter, set on the basis of the nature of the research being undertaken.e) in symbolic terms, the total signal read out from a pixel after an exposure of length t taking into account the quasar flux F (per unit time), the dark current D (per unit time), and the bias B.

TotalSignalReadout=F-D-B

TSR=((1*109photons/second)*90%)-(2electrons/second/pixel)-2000electrons

For the total signal readout you need the electrons coming in per second (F), or “flux” subtracting the noise, or Dark images/frames and the bias.f)the limiting exposure time, after setting the total signal read out to be the identified maximum.TotalSignalReadout=F-D-B

TSR=((1*109photons/second)*90%)-(2electrons/second/pixel)-2000electrons

*Solving for s (second)*

TotalSignalReadout=F-D-B

Max=((1*109photons/second)*90%)-(2electrons/second/pixel)-2000electrons

84000=((1*109photons/second)*90%)-(2electrons/second/pixel)-2000electrons

s=10465.11 OR 10465

s/60=minutes

minutes=10465s/60s

minutes=174.25

Here i used the total signal read out equation, subbed in the maximum amount of electrons obsorbed to solve for s (seconds), and then convert into minutes.CONCLUSION:The maximum amount of time, in minutes, that you should keep the shutter open is 174.25 minutes to get an accurate CCD picture of a quasar flux that is the 16th magnitude.I just need to know if what i came up with makes sort of sense?? I know its long, and I am sorry for that. This was the first question given to me after ONE intro class so I am completely lost. I don't know if I'm just bad at physics or if my prof has overestimated what we knew in review.

Thank you so much for any help!