Use the inverse square law for light to answer each of the...

Use the inverse square law for light to answer each of the following questions. a) Suppose a star has the same luminosity as our Sun (3.8×10^26 watts) but is located at a distance of 10 light-years. What is its apparent brightness? Express your answer using two significant figures. b) Suppose a star has the same apparent brightness as Alpha Centauri A (2.7×10^-8 watt/m^2) but is located at a distance of 300 light-years. What is its luminosity? Express your answer using two significant figures. c) Suppose a star has a luminosity of 7.0×10^26 watts and an apparent brightness of 3.0×10^-12 watt/m^2. How far away is it? Give your answer in both kilometers and light-years. Express your answer using two significant figures.

Transcript

00:01 Hello everyone here the question consists three parts part a part b part c so now coming to part a so in part a it is given that luminosity l equals to 3 .8 multiplied by 10 to the power 26 watt and distance b equals to given 10 light years which we can attain as 9 .461 multiplied by 10 to the power 15 multiplied by 10 meter so we will convert the lighters into meters so now in this case we have to find the apparent bright end the value of f so now as we know that apparent by 10 is given by the formula f equals to l divided by 4 pi b squared so now here we will put the values so therefore from here we will get f equals to 3 .8 multiplied by 10 to the power 26 divided by 4 pi multiplied by d squared the value of d squared we already given so the value of d square is 9 .461 multiplied by 10 multiplied by 10 to the over 15 whole square so therefore further solving this from here we will get apparent brightness f equals to 3 .33 multiplied by 10 to the power minus 9 watt part meter square so now coming to the next part next part means part p in part b of the position it is given that apparent brightness f equals to 2 .7 multiplied by 10 3 3d000 1st 8 watt part meter squared and distance d equals to given 300 lightest which we will convert into meter so therefore we will get d equals to 300 multiplied by 9 .461 multiplied by 1030 over 15 meter so here in this case we have to find the luminosity so again we are using the above formula so using the formula from here we will get luminosity l equals to f multiplied by 4 pi b squared so now here we will put the values so therefore 2 .7 multiplied by 103 minus 8 multiplied by 4 multiplied by 4 multiplied by pi multiplied by d square the value of this square is 300 multiplied by 10 to the 15 multiplied by 9 .461 whole square so therefore further solving this we will get luminosity l equals to 2 .7 multiplied by 10 to over 301 so now coming to the last part which is part c in part c of the question it is given that luminosity l equals to 7 multiplied by 10 to the power 26 watt and apparent brightness f equals to given 3 multiplied by 10 0 minus 12 watt per meter squared so therefore here in this case we have to find the distance so now using the same formula from here we will get distance b equals to square root of l divided by f multiplied by 4 pi so now here we will put the values so therefore d equals to square root of 7 multiplied by 10 to the power 26 divided by the value of f is 3 multiplied by 10 to 2 minus 12 multiplied by 4 multiplied by 5.

03:55 So therefore further solving this from here we will get distance d equals to 4 .3 multiplied by 10 to 10 to the over 18 meter.

04:05 So now distance in lightest we have to divide by 9 .461 multiplied by 30 by 50...

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