Imaging from the Earth's surface causes a problem when even on a clear night the atmosphere emits a faint glow, equivalent to a V
21 star everywhere you look.On such a night, how long would you need to observe on the a 8.2m telescope to obtain an optical spectrum of a V
20 magnitude elliptical galaxy where the signal to noise ratio per nanometer is 25 ?Assume the spectrograph spreads the photons received evenly across the V-filter bandpass. SOLVE IT AND GIVE ANSWERS
To solve this problem, we can use the formula for signal-to-noise ratio (SNR):
SNR = (Signal / Noise)
We are given that the SNR per nanometer is 25. The signal is the number of photons received from the galaxy within the V-filter bandpass, and the noise is the faint glow emitted by the atmosphere.
First, let's calculate the signal:
We know that the magnitude difference between the galaxy and the faint glow is 20 - (-21) = 41 magnitudes. The magnitude system is logarithmic, so the flux ratio between the galaxy and the faint glow is given by:
Flux ratio = 10^(-0.4 * (V_star - V_galaxy))
Since V_star is -21 and V_galaxy is 20, the flux ratio is:
Flux ratio = 10^(-0.4 * (-21 - 20))
= 10^(-0.4 * (-41))
= 10^(16.4)
≈ 2.5 * 10^16
The total signal received from the galaxy within the V-filter bandpass is directly proportional to the flux ratio. Let's assume there are N photons received from the galaxy within the V-filter bandpass:
N ∝ Flux ratio
N = k * Flux ratio
We can solve for k by using the fact that the V-filter has a standard magnitude zero point, which is roughly 3640 photons/cm^2/s/Angstrom. This means that if a star of magnitude 0 is observed through the V-filter, it would produce 3640 photons/cm^2/s/Angstrom:
k * Flux ratio = 3640
Substituting the value of Flux ratio, we get:
k * (2.5 * 10^16) = 3640
k ≈ 3640 / (2.5 * 10^16)
≈ 1.45 * 10^(-13)
Now, we can calculate the signal received from the galaxy within the V-filter bandpass:
Signal = k * Flux ratio
≈ (1.45 * 10^(-13)) * (2.5 * 10^16)
≈ 3.63
Next, let's calculate the noise:
The noise emitted by the atmosphere is equivalent to a V-band magnitude of -21. Assuming the same V-filter zero point, the noise would produce the following number of photons:
Noise = -21 * 3640
≈ -76440
The noise is negative because it represents a faint glow, but we will take its absolute value as it does not affect the calculation of the SNR.
Finally, we can calculate the SNR:
SNR = Signal / Noise
= 3.63 / |(-76440)|
≈ 4.76 * 10^(-5)
To obtain a reliable spectrum, astronomers generally aim for an SNR of at least 10. Let's assume we want an SNR of 10:
SNR = 10 = Signal / Noise
Signal = 10 * Noise
= 10 * |(-76440)|
= 764400
Now, let's calculate the time required to collect enough photons to achieve this signal:
Assuming the telescope collects all the photons from the galaxy within the V-filter bandpass, the number of photons collected per second can be calculated using the formula:
Number of photons per second = Telescope collecting area * Signal
The collecting area of the 8.2m telescope can be calculated using the formula for the area of a circle:
Area = π * (Radius)^2
Given a telescope diameter of 8.2m, the radius can be calculated as:
Radius = Diameter / 2
= 8.2 / 2
= 4.1m
Substituting the value of the radius, we can calculate the collecting area:
Area = π * (4.1)^2
≈ 53.14 m^2
Now, let's calculate the number of photons collected per second:
Number of photons per second = 53.14 * 764400
Assuming the photons are collected evenly across the V-filter bandpass, the number of photons collected per nanometer can be calculated using the formula:
Number of photons per nanometer = Number of photons per second / Bandpass
The V-filter bandpass is typically around 90 nm. Therefore:
Number of photons per nanometer = (53.14 * 764400) / 90
Finally, we can calculate the time required to obtain the optical spectrum by finding the reciprocal of the number of photons collected per nanometer:
Time = 1 / (Number of photons per nanometer)
Solving this equation will give us the desired time in seconds:
Time = 1 / [(53.14 * 764400) / 90]
Calculating this equation will give us the final answer for the time required to obtain the optical spectrum.
To calculate the total observing time needed to obtain an optical spectrum of a V
20 magnitude elliptical galaxy in an 8.2m telescope, we need to consider the signal-to-noise ratio and the photon flux.
Step 1: Calculate the photon flux
The photon flux per nanometer can be obtained using the formula:
Photon flux = 10^(-0.4 * (V + 21))
Where V is the magnitude of the galaxy we want to observe. In this case, it is V
20.
Photon flux = 10^(-0.4 * (20 + 21))
Photon flux = 10^(-0.4 * 41)
Photon flux = 10^(-16.4)
Step 2: Calculate the signal-to-noise ratio (SNR) per nanometer
Given that the SNR per nanometer is 25, we can use the formula:
SNR = sqrt(N_photons) / sqrt(N_background)
Where N_photons is the number of photons from the source and N_background is the number of background photons.
Since the spectrograph spreads the photons evenly across the V-filter bandpass, we can assume the background is also evenly spread. Thus, the background photon count will be equal to the sum of the photon flux from the atmosphere and other sources.
N_background = Photon flux + Background_flux
N_background = 10^(-16.4) + B
Where B is the background flux. For simplicity, we will assume the background flux is approximately equal to the photon flux.
N_background = 2 * 10^(-16.4)
Therefore, the SNR can be calculated as follows:
25 = sqrt(N_photons) / sqrt(2 * 10^(-16.4))
Step 3: Solve for N_photons
To solve for N_photons, we rearrange the equation as follows:
N_photons = (25^2) * 2 * 10^(-16.4)
N_photons = 625 * 2 * 10^(-16.4)
N_photons = 1250 * 10^(-16.4)
Step 4: Calculate the exposure time (t_exp)
To calculate the exposure time, we can use the formula:
t_exp = N_photons / Photon_flux
t_exp = (1250 * 10^(-16.4)) / 10^(-16.4)
t_exp = 1250 seconds
Therefore, you would need to observe for approximately 1250 seconds or about 20.8 minutes on the 8.2m telescope to obtain an optical spectrum of a V
20 magnitude elliptical galaxy with a signal-to-noise ratio of 25 per nanometer.
To calculate the time needed to obtain an optical spectrum of a galaxy with a specific signal-to-noise ratio, we can use the formula:
t = (S/N)^2 × (F × t_exp)/(R × B)
Where:
t = time needed to observe (in seconds)
S/N = signal to noise ratio
F = flux (photons per square meter per second) from the galaxy
t_exp = exposure time (in seconds)
R = telescope's spectral resolution (in nm per pixel)
B = bandpass width (in nm)
We are given:
S/N = 25
F = flux from the galaxy
R = 8.2m telescope (spectral resolution needed to be determined)
B = width of the V-filter bandpass (to be determined)
First, we need to determine the flux from the galaxy. In the question, it states that the galaxy has a magnitude of V = 20. To convert this to flux, we can use the formula:
F = F_v0 × 10^(-0.4 × (V - V0))
Where:
F_v0 = flux of a reference magnitude V0
V0 = reference magnitude (often taken as zero, so F_v0 = 3631 Jy)
Now we can calculate the flux:
F = 3631 Jy × 10^(-0.4 × (20 - 0))
Next, we need to determine the spectral resolution (R) and the bandpass width (B) for the V-filter.
The standard R for a typical spectrograph can be calculated using the formula:
R = λ / Δλ
Where λ is the central wavelength of the spectrum and Δλ is the full-width at half-maximum (FWHM) of the spectral line.
For the V-filter, the central wavelength is approximately 550 nm, and the FWHM can be taken as 90 nm.
Finally, we can substitute all the values into the formula to solve for the exposure time (t_exp):
t_exp = (t × R × B) / (S/N)^2 × F
Note: The values for F, R, and B need to be converted into appropriate units (e.g., photons per square meter per second, nm per pixel) to ensure consistency in the calculation.
Plug in the values, and solve the equation to find the exposure time required for observing the galaxy with the desired signal-to-noise ratio.