Recent content by lquinnl

Yeah that makes sense now! Thank you very much for sticking with me and helping out with this problem. I'm trying to revise for an exam and I was hittting a brick wall with that! Thanks again

So the ratio of this F_{\lambda}(T) = \frac{2h\pi c^{2}}{\lambda^{5}} \frac{1}{exp(hc/\lambda KT)-1} Wm^{-2}m^{-1} to this: F_\lambda = \frac{\Delta E}{\Delta A \Delta t \Delta \lambda} is (\frac{R}{d})^{2}

Ooops so 650nm should be 6.5 x10^-7 m. So compare F_{\lambda}(T) = \frac{2h\pi c^{2}}{\lambda^{5}} \frac{1}{exp(hc/\lambda KT)-1} Wm^{2}m^{-1} with F_{\lambda}(T) = \frac{\Delta E}{\Delta A \Delta \lambda \Delta t} then use this ratio, call it Z ,to calculate the Fbolometric at...

OK so I have: Monochromatic flux detected from Earth: F_{\lambda} = \frac{\Delta E}{\Delta A \Delta \lambda \Delta t} The measured spectrum of the star peaks at a wavelength of 650 nm. Assuming the star radiates as a black body, T = \frac{2.898 \times 10^{-3} mK}{6.5\times 10^{-9}m}...

A star X is observed in the V-band filter (central wavelength 550 nm, bandwidth 88 nm) using a telescope with a diameter of 40cm. The telescope and camera detect 50% of the incident photons and, during a 10 sec exposure, 5500 photons are detected. I have calculated the monochromatic flux of the...

I know the apparent magnitude is related to flux by the relationship m=-2.5log_{10}(F)+ constant am i to somehow use this to calculate the bolometric flux at the given distance then use this flux on the left hand side of F^{bol} = (\frac{R}{d})^{2}\sigma T^{4} Although i do not know...

Homework Statement The monochromatic flux emitted from unit surface area of a black body is given by F_{\lambda}(T) = \frac{2h\pi c^{2}}{\lambda^{5}} \frac{1}{exp(hc/\lambda KT)-1} Wm^{-2}m^{-1} If the distance to star X is 620 parsecs, calculate: (a) the radius of star X, in...

THANK YOU VERY MUCH! I feel like I have done an incredible job of over-complicating this problem! I think I was getting confused by the fact we have F_v and F_lambda. So it was just a case of expressing wavelength as frequency and amalgamating the powers. Seems incredibly straightforward...

I thought it should be minus 2 from this statement: F_v \propto v^{\beta} \propto (\lambda^{-1})\propto(v^{2})^{-1}

So if F_v \propto v^{\beta} \propto v^{-2} \beta = -2

What I'm saying is that F_\lambda \propto \lambda^{-2} \propto v^{2} from v = \frac{c}{\lambda} we have v \propto \lambda^{-1} so F_v \propto \lambda^{-1} now i think maybe this should be how i follow:so F_v \propto v^{\beta} \propto (\lambda^{-1})\propto(v^{2})^{-1}

Or have I messed up still, regarding a sign?

But (\nu^{-1})^{-2} is NOT \nu^{-3}. I note that you made a similar mistake in another derivation. What is the proper result of such an operation? OF COURSE! I feel so stupid today, really need to wake up! (v^{-1})^{-2} = v^{2} so F_v \propto v^{\beta} \propto \lambda^{-1} \propto...

F_\lambda \propto \lambda^{-2} \propto (v^{-1})^{-2} \propto v^{-3} v \propto \lambda^{-1} F_v \propto v^{\beta} \propto \lambda^{-1} \propto (v^{-3})^{-1} \propto v^{-4} so \beta = 4 ? P.s. Thanks for all this help.

Looking at this again I am not convinced of my reasoning. I have this: F_\lambda \propto \lambda^{-2} . F_v \propto v^{-\beta} . F_v \propto v^{-\beta} \propto {\frac{c}{\lambda}}^{-\beta} . Any chance of a poke in the right direction, if you know?