1) the explosion occurs [tex]5.83\cdot 10^8[/tex] years ago
2) [tex]4.25\cdot 10^{14} Hz[/tex]
Explanation:
1)
The light coming from the exploding supernova travels at constant speed, which is the speed of light:
[tex]c=3.0\cdot 10^8 m/s[/tex]
And since it travels with uniform motion, the distance it covers in a time t is given by:
[tex]d=ct[/tex]
where t is the time taken.
Here, we know that the distance of the supernova from the Earth is
[tex]d=5.83\cdot 10^8 ly[/tex]
where 1 light-year is the distance covered by light in 1 year. This means that in 1 year, light covers a distance of 1 light-year; in 5 year, light covers a distance of 5 light-years; and so on.
Therefore here, since the distance is
[tex]d=5.83\cdot 10^8 ly[/tex]
This means that the time taken by light is
[tex]t=5.83\cdot 10^8[/tex] years
And therefore, the explosion occurs [tex]5.83\cdot 10^8[/tex] years ago.
2)
First of all, the real frequency of the light emitted by the supernova is given by
[tex]f=\frac{c}{\lambda}[/tex]
where
c is the speed of light
[tex]\lambda=685 nm = 685\cdot 10^{-9}m[/tex] is the proper wavelength
Substituting,
[tex]f=\frac{3.0\cdot 10^8}{685\cdot 10^{-9}}=4.38\cdot 10^{14} Hz[/tex]
The frequency observed on Earth is shifted (Doppler effect) due to the relative motion between the supernova and the Earth; the apparent frequency observed on Earth is given by
[tex]f'=\frac{c}{c-v}f[/tex]
where
c is the speed of light
[tex]v=8.85\cdot 10^6 m/s[/tex] is the speed at which the supernova is moving apart
Substituting the values, we find the apparent frequency:
[tex]f'=\frac{3\cdot 10^8}{3\cdot 10^8+8.85\cdot 10^6}(4.38\cdot 10^{14})=4.25\cdot 10^{14} Hz[/tex]
