<p>The F850LP has a mean wavelength (filter + CCD response) of 9193 angstroms and <a href="http://svo2.cab.inta-csic.es/svo/theory/fps3/index.php?id=HST/ACS_HRC.F850LP" rel="nofollow noreferrer">this appears to be</a> the reddest filter that can be used with the ACS instrument. Optimal for observing and identifying high redshift sources and at a distinct central wavelength to all the other filters<span class="math-container">$^1$</span>. This latter property is important to gain any sort of spectral resolution from this kind of photometric redshift technique and hence a better handle on the source redshift.</p>
<p>The &quot;LP&quot; in the filter designation means it is a &quot;long-pass&quot; filter, in contrast to the other filters shown in the plot. That means that what Dr Becky's graph is showing for F850LP is NOT the filter transmission curve, it is the total response curve of the filter and the CCD. The optical CCD response rolls off smoothly above 900 nm and that is what defines the red side of the response curve, not the filter.</p>
<p>However, the plot is doubly confusing because it does appear to be the filter transmissions that are shown for the other filters rather than the product of the filter transmission and CCD response. A summary table of the ACS filter (only) transmissions are given <a href="https://etc.stsci.edu/etcstatic/users_guide/appendix_b_acs.html" rel="nofollow noreferrer">here</a> and you can see that the FWHM of the F850LP filter transmission is unbounded because there is no red cut-off.</p>
<p>The F850LP is part of <a href="https://hst-docs.stsci.edu/acsihb/chapter-5-imaging/5-3-wide-field-optical-ccd-imaging" rel="nofollow noreferrer">the set</a> F475W, F625W, F775W, F850LP that approximate to the Sloan g, r, i, z filters.</p>
<p><span class="math-container">$^1$</span> The <a href="https://stsci-opo.org/STScI-01FX61F3NBSQQY4K41GKAX05QD.pdf" rel="nofollow noreferrer">draft paper</a> you pointed to in comments suggests that the object was <strong>not</strong> observed with the F850LP filter at all!</p>

CC BY-SA 4.0
2022-04-09T11:20:50.233Z
Interesting! Won't be able to see the [Nature paper](https://www.nature.com/articles/s41586-022-04449-y) until Monday (no arXiv preprint?) but I just found out that under Dr. Becky's video the notes link to [this draft](https://stsci-opo.org/STScI-01FX61F3NBSQQY4K41GKAX05QD.pdf) and in Extended Data Table 1. (page 42) it doesn't even show F850LP, though it does show that F814W is used with the HST Advanced Camera for Surveys (ACS) which should have the silicon cutoff [here](http://svo2.cab.inta-csic.es/svo/theory/fps3/index.php?id=HST/ACS_HRC.F814W&&mode=browse&gname=HST&gname2=ACS_HRC#filter)
uhoh
CC BY-SA 4.0
2022-04-09T11:20:58.680Z
but the plot in the video shows it as flat to about 0.95, suggesting ACSs silicon CCD response is not folded in for this filter in the plot. I can't find any data for F814W for the [WFC3_IR](http://svo2.cab.inta-csic.es/svo/theory/fps3/index.php?mode=browse&gname=HST&gname2=WFC3_IR&asttype=) and so there may be remaining ambiguities in the plot in the video, but it certainly doesn't affect your answer.
uhoh
CC BY-SA 4.0
2022-04-09T11:40:51.847Z
The red cut-off of F814W is defined by the filter at 960nm. I agree though that  it appears to be the filter throughput shown in the plot. The whole figure is confused/confusing. @uhoh
ProfRob
CC BY-SA 4.0
2022-04-10T16:47:00.323Z
Trivial nitpick: “LP” = “long pass”, not “low pass”.
Peter Erwin
CC BY-SA 4.0
2022-04-10T16:56:30.693Z
@PeterErwin I guess so. One of the web pages above referred to them as low-pass, which I then thought referred to the frequency. But I think you are right.
ProfRob
CC BY-SA 4.0
2022-04-09T10:37:15.057Z
2022-04-10T16:55:10.460Z
2022-04-10T16:55:10.460Z
ProfRob
<p>It will take some time to read through <a href="https://astronomy.stackexchange.com/a/48996/7982">@Pela's excellent answer to <em>Is there a better explanation of this picture showing the very distant star &quot;Earendel&quot;?</em></a> and to slowly ease myself into it I've just watched the new Dr. Becky video <a href="https://www.youtube.com/watch?v=VChgsXbIgdw" rel="noreferrer">ALL THE DETAILS on Earendel: a star 12.9 billion light years away seen by the Hubble Space Telescope</a>.</p>
<p>She explains that <a href="https://en.wikipedia.org/wiki/WHL0137-LS" rel="noreferrer">WHL0137-LS</a> is pretty dim and so attempts to determine if it is as red-shifted (and therefore old) as the rest of the lensed bits using Hubble's spectroscopy would be challenging, so instead images through a variety of filters can be used to at least try to rule out the possibility of it being a foreground start rather than a distant star lensed and magnified in brightness by a factor of at least thousands.</p>
<p>I noticed in the graphic (shown in the screenshot below) which illustrates all the filters used in this study, that while all but one have conventional square-ish bandpass shapes with fairly straight sides and roughly flat tops (except for the inevitable little wiggles across the top in the transmission region). The exception is the filter F850LP, which has a very differently-shaped transmission function. It has a narrower peak bandpas and a much more gradual fall-off on the long-wavelength side, sort-of a rounded sawtooth shape.</p>
<p><strong>Question:</strong> Why does Hubble's F850LP filter have a different shape than all the others shown in Dr. Becky's video &quot;ALL THE DETAILS on Earendel...&quot;?</p>
<p>I think an answer can address both</p>
<ol>
<li>Why this filter F850LP has a different shape than all the rest (e.g. technical limitations, legacy (consistency with historical data), something else)?</li>
<li>Why was it included in this study? This likely took quite a lot of observing time, and since each filter requires a new exposure filter choices would have had some careful consideration.</li>
</ol>
<hr />
<p>Screenshot from Dr. Becky video <a href="https://www.youtube.com/watch?v=VChgsXbIgdw" rel="noreferrer">ALL THE DETAILS on Earendel: a star 12.9 billion light years away seen by the Hubble Space Telescope</a> showing the transmission response of all the HST filters used in the study:</p>
<p><a href="https://i.sstatic.net/0rSkT.jpg" rel="noreferrer"><img src="https://i.sstatic.net/0rSkT.jpg" alt="Screenshot from Dr. Becky video &quot;ALL THE DETAILS on Earendel: a star 12.9 billion light years away seen by the Hubble Space Telescope&quot; https://www.youtube.com/watch?v=VChgsXbIgdw showing the transmission response of all the HST filters used in the study" /></a></p>

<observational-astronomy><spectroscopy><hubble-telescope><spectrometry><earendel-whl0137-ls>
Why does Hubble's F850LP filter have a different shape than all the others shown in Dr. Becky's video "ALL THE DETAILS on Earendel..."?
uhoh
# Why does Hubble's F850LP filter have a different shape than all the others shown in Dr. Becky's video "ALL THE DETAILS on Earendel..."?

(Asked by: username_0 on 2022-04-08)

It will take some time to read through [@Pela's excellent answer to *Is there a better explanation of this picture showing the very distant star "Earendel"?*](https://astronomy.stackexchange.com/a/48996/7982) and to slowly ease myself into it I've just watched the new Dr. Becky video [ALL THE DETAILS on Earendel: a star 12.9 billion light years away seen by the Hubble Space Telescope](https://www.youtube.com/watch?v=VChgsXbIgdw).

She explains that [WHL0137-LS](https://en.wikipedia.org/wiki/WHL0137-LS) is pretty dim and so attempts to determine if it is as red-shifted (and therefore old) as the rest of the lensed bits using Hubble's spectroscopy would be challenging, so instead images through a variety of filters can be used to at least try to rule out the possibility of it being a foreground start rather than a distant star lensed and magnified in brightness by a factor of at least thousands.

I noticed in the graphic (shown in the screenshot below) which illustrates all the filters used in this study, that while all but one have conventional square-ish bandpass shapes with fairly straight sides and roughly flat tops (except for the inevitable little wiggles across the top in the transmission region). The exception is the filter F850LP, which has a very differently-shaped transmission function. It has a narrower peak bandpas and a much more gradual fall-off on the long-wavelength side, sort-of a rounded sawtooth shape.

**Question:** Why does Hubble's F850LP filter have a different shape than all the others shown in Dr. Becky's video "ALL THE DETAILS on Earendel..."?

I think an answer can address both

1. Why this filter F850LP has a different shape than all the rest (e.g. technical limitations, legacy (consistency with historical data), something else)?
2. Why was it included in this study? This likely took quite a lot of observing time, and since each filter requires a new exposure filter choices would have had some careful consideration.

---

Screenshot from Dr. Becky video [ALL THE DETAILS on Earendel: a star 12.9 billion light years away seen by the Hubble Space Telescope](https://www.youtube.com/watch?v=VChgsXbIgdw) showing the transmission response of all the HST filters used in the study:

[![Screenshot from Dr. Becky video ](https://i.sstatic.net/0rSkT.jpg)](https://i.sstatic.net/0rSkT.jpg)

**Tags:** observational-astronomy, spectroscopy, hubble-telescope, spectrometry, earendel-whl0137-ls

---

## 1 answers

### Answer by username_1 on 2022-04-09. Score: 2 | Accepted answer

The F850LP has a mean wavelength (filter + CCD response) of 9193 angstroms and [this appears to be](http://svo2.cab.inta-csic.es/svo/theory/fps3/index.php?id=HST/ACS_HRC.F850LP) the reddest filter that can be used with the ACS instrument. Optimal for observing and identifying high redshift sources and at a distinct central wavelength to all the other filters$^1$. This latter property is important to gain any sort of spectral resolution from this kind of photometric redshift technique and hence a better handle on the source redshift.

The "LP" in the filter designation means it is a "long-pass" filter, in contrast to the other filters shown in the plot. That means that what Dr Becky's graph is showing for F850LP is NOT the filter transmission curve, it is the total response curve of the filter and the CCD. The optical CCD response rolls off smoothly above 900 nm and that is what defines the red side of the response curve, not the filter.

However, the plot is doubly confusing because it does appear to be the filter transmissions that are shown for the other filters rather than the product of the filter transmission and CCD response. A summary table of the ACS filter (only) transmissions are given [here](https://etc.stsci.edu/etcstatic/users_guide/appendix_b_acs.html) and you can see that the FWHM of the F850LP filter transmission is unbounded because there is no red cut-off.

The F850LP is part of [the set](https://hst-docs.stsci.edu/acsihb/chapter-5-imaging/5-3-wide-field-optical-ccd-imaging) F475W, F625W, F775W, F850LP that approximate to the Sloan g, r, i, z filters.

$^1$ The [draft paper](https://stsci-opo.org/STScI-01FX61F3NBSQQY4K41GKAX05QD.pdf) you pointed to in comments suggests that the object was **not** observed with the F850LP filter at all!

> **Comments:**
> **username_0**: Interesting! Won't be able to see the [Nature paper](https://www.nature.com/articles/s41586-022-04449-y) until Monday (no arXiv preprint?) but I just found out that under Dr. Becky's video the notes link to [this draft](https://stsci-opo.org/STScI-01FX61F3NBSQQY4K41GKAX05QD.pdf) and in Extended Data Table 1. (page 42) it doesn't even show F850LP, though it does show that F814W is used with the HST Advanced Camera for Surveys (ACS) which should have the silicon cutoff [here](http://svo2.cab.inta-csic.es/svo/theory/fps3/index.php?id=HST/ACS_HRC.F814W&&mode=browse&gname=HST&gname2=ACS_HRC#filter)  
> **username_0**: but the plot in the video shows it as flat to about 0.95, suggesting ACSs silicon CCD response is not folded in for this filter in the plot. I can't find any data for F814W for the [WFC3_IR](http://svo2.cab.inta-csic.es/svo/theory/fps3/index.php?mode=browse&gname=HST&gname2=WFC3_IR&asttype=) and so there may be remaining ambiguities in the plot in the video, but it certainly doesn't affect your answer.  
> **username_1**: The red cut-off of F814W is defined by the filter at 960nm. I agree though that  it appears to be the filter throughput shown in the plot. The whole figure is confused/confusing. @uhoh  
> **username_2**: Trivial nitpick: “LP” = “long pass”, not “low pass”.  
> **username_1**: @PeterErwin I guess so. One of the web pages above referred to them as low-pass, which I then thought referred to the frequency. But I think you are right.  

---

