In 1987, a supernova was observed, so it was 34 years ago (2021). The scientist believed that galaxy was $d= 1.66 cdot 10^{21} hspace{0.5mm} mathrm{m}$ away from us. In 1987, they observed the light from that galaxy, and the speed of that light was $c = 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}$. To calculate the time for which the light was travel, we use the equation [ d= ct] Now, we can combine these information to calculate the time
begin{align*}
d&=ct\
t&= frac{d}{c}\
t&= frac{1.66 cdot 10^{21} hspace{0.5mm} mathrm{m}}{ 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}}\
t&=5.53 cdot 10^{12} hspace{0.5mm} mathrm{s}\
t&=5.53 cdot 10^{12} hspace{0.5mm} mathrm{s} cdot mathrm{frac{1 yr}{ 31.536 cdot 10^{6} s}} \
t&=1.75 cdot 10^{5} hspace{0.5mm} mathrm{yr}
end{align*}
To get precise value of time, we should add 34 years, but it is neglible to the value of $t$. So, we can say that te supernova explosion actually occur before [ framebox[1.1width]{$ therefore t=1.75 cdot 10^{5} hspace{0.5mm} mathrm{yr} $}]

$B.$ $1.75 cdot 10^{5} hspace{0.5mm} mathrm{yr}$

The speed of a galaxy is $v=5.8 cdot 10^{6} hspace{0.5mm} mathrm{m/s}$, and it is moving away from us. An observer sees the light from that galaxy with a frequency of $f_{obs}=5.6 cdot 10^{14} hspace{0.5mm} mathrm{Hz}$. To find the frequency of emitted light $f$, we need to use equation [f_{obs}=f(1- frac{v}{c})] where $c= 3 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}$. Now, we can calculate the frequency of emitted ligth
begin{align*}
f_{obs}&=f(1- frac{v}{c}) \
f&=f_{obs} (1- frac{v}{c})^{-1} \
f&= 5.6 cdot 10^{14} hspace{0.5mm} mathrm{Hz} (1- frac{5.8 cdot 10^{6} hspace{0.5mm} mathrm{m/s}}{3 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}})^{-1} \
f&= 5.6 cdot 10^{14} hspace{0.5mm} mathrm{Hz} (1- 1.93 cdot 10^{-2})^{-1} \
f&= frac{5.6}{0.98} cdot 10^{14} hspace{0.5mm} mathrm{Hz} \
f&= 5.7 cdot 10^{14} hspace{0.5mm} mathrm{Hz}
end{align*}
The frequency of emitted light is [ framebox[1.1width]{$ therefore f= 5.7 cdot 10^{14} hspace{0.5mm} mathrm{Hz} $}]

$C.$ $f= 5.7 cdot 10^{14} hspace{0.5mm} mathrm{Hz}$

The illuminence of the sunlight is $E= 1 cdot 10^{5} hspace{0.5mm} mathrm{lx}$. A light on a stage has the luminous intensity of $5 cdot 10^{6} hspace{0.5mm} mathrm{cd}$, or if we write this as the luminous flux $P=4 pi 5 cdot 10^{6} hspace{0.5mm} mathrm{lm}$. To find the distance from a stage, we use the equation [E= frac{P}{4 pi d^{2}}] where $d$ is the distance from a stage. Now, we can calculate the distance as
begin{align*}
E&= frac{P}{4 pi d^{2}} \
d^{2}&= frac{P}{4 pi E} \
d^{2}&= frac{ 4 pi 5 cdot 10^{6} hspace{0.5mm} mathrm{lm}}{4 pi cdot 1 cdot 10^{5} hspace{0.5mm} mathrm{lx} } \
d^{2}&= 50 hspace{0.5mm} mathrm{m^{2}} \
d&= 7.1 hspace{0.5mm} mathrm{m}
end{align*}
A member of audience should be at the distance of [ framebox[1.1width]{$ therefore d= 7.1 hspace{0.5mm} mathrm{m} $}]

$B.$ $d= 7.1 hspace{0.5mm} mathrm{m}$

The illuminance of $60.0 hspace{0.5mm} mathrm{W}$ lightbulb is $E=9.35 hspace{0.5mm} mathrm{lx}$. Also, we know that lightbulb is at the distance of $d= 3.0 hspace{0.5mm} mathrm{m}$ from illuminated surface. To calculate the total luminous flux of the bulb we us [E= frac{P}{4 pi d^{2}}] where $P$ is the total luminous flux.Now, we can calculate the luminous flux as
begin{align*}
E&= frac{P}{4 pi d^{2}} \
P&= 4 pi E d^{2}\
P&= 4 pi 9.35 hspace{0.5mm} mathrm{lx} cdot (3.0 hspace{0.5mm} mathrm{m})^{2}\
P&=1057.46 hspace{0.5mm} mathrm{lm}\
P&=1.1 cdot 10^{3} hspace{0.5mm} mathrm{lm}
end{align*}
The total luminous flux of the bulb is [ framebox[1.1width]{$ therefore P=1.1 cdot 10^{3} hspace{0.5mm} mathrm{lm} $}]

$D.$ $P=1.1 cdot 10^{3} hspace{0.5mm} mathrm{lm}$

The light from the Sun travels through space, so the light travels in a vacuum, with the speed $c = 3 cdot 10^{8} hspace{0.5mm} mathrm{m/s}$. The sunlight travels for [t= 8.0 hspace{0.5mm} mathrm{min} Rightarrow t= 480.0 hspace{0.5mm} mathrm{s}] so we can use the equation [d= ct] Now, we can calculate the distance
begin{align*}
d&= ct\
d&= 3 cdot 10^{8} hspace{0.5mm} mathrm{m/s} cdot 480.0 hspace{0.5mm} mathrm{s} \
d&= 1440 cdot 10^{8} hspace{0.5mm} mathrm{m} \
d&= 1.4 cdot 10^{11} hspace{0.5mm} mathrm{m} \
d&= 1.4 cdot 10^{8} hspace{0.5mm} mathrm{km}
end{align*}
The distance between the Sun and Earth is [ framebox[1.1width]{$ therefore d= 1.4 cdot 10^{8} hspace{0.5mm} mathrm{km} $}]

$C.$ $1.4 cdot 10^{8} hspace{0.5mm} mathrm{km}$

The speed of light in a vacuum is $c= 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}$. The light has a wavelength of [ lambda=404 hspace{0.5mm} mathrm{nm} Rightarrow lambda=404 cdot 10^{-9} hspace{0.5mm} mathrm{m} ] Now, we can calculate the frequency
begin{align*}
f&= frac{c}{ lambda} \
f&= frac{ 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}}{ 404 cdot 10^{-9} hspace{0.5mm} mathrm{m}} \
f&= 7.43 cdot 10^{14} hspace{0.5mm} mathrm{Hz}
end{align*}
The frequency of the light is [ framebox[1.1width]{$ therefore f= 7.43 cdot 10^{14} hspace{0.5mm} mathrm{Hz} $}]

$D.$ $7.43 cdot 10^{14} hspace{0.5mm} mathrm{Hz}$

The wavelength of emitted light is [ lambda = 525 hspace{0.5mm} mathrm{nm} ] The observed wavelength of light is [ lambda_{obs} = 473 hspace{0.5mm} mathrm{nm} ] Now, we can calculate the speed of an object
begin{align*}
lambda_{obs} – lambda &= pm frac{v}{c} lambda / : lambda \
frac{lambda_{obs}}{lambda} – 1 &=pm frac{v}{c}\
pm frac{v}{c}&= frac{ 473 hspace{0.5mm} mathrm{nm} }{ 525 hspace{0.5mm} mathrm{nm}} – 1 \
pm frac{v}{c}&= frac{473}{525} – 1 \
pm frac{v}{c}&= -0.099
end{align*}
Part of the equation $ frac{v}{c}$ has to be greater than zero, so the sign is “-“, which means, the celectial object is approaching.
begin{align*}
– frac{v}{c}&= -0.099\
frac{v}{c}&= 0.099\
v&= 0.099 c\
v&= 0.099 cdot 3 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}\
v&= 0.297 cdot 10^{8} hspace{0.5mm} mathrm{frac{m}{s}}\
v&= 29.7 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}\
end{align*}
The speed of a celectial object is [ framebox[1.1width]{$ therefore v= 29.7 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}} $}]

$$
v= 29.7 cdot 10^{6} hspace{0.5mm} mathrm{frac{m}{s}}
$$

Intensity of nonpolarized light is $I_{0}$. That nonpolarized light goes through the polarizing filter, and the transmitted light is now polarized , with the intensity of [I_{1}= frac{I_{0}}{2} ] When polarized light goes through the polarizing filter, we can apply Malus’s Law [ I_{2} = I_{1} cos^{2} alpha ] In our case, the angle is $ alpha = 45 textdegree$, so we can calculate the intensity of transmitted light $I_{2}$
begin{align*}
I_{1}&= frac{I_{0}}{2}\
I_{2}&= I_{1} cos^{2} alpha \
I_{2}&= frac{I_{0}}{2} cos^{2} alpha \
I_{2}&= frac{I_{0}}{2} cos^{2} 45 textdegree \
I_{2}&= frac{I_{0}}{2} ( frac{ sqrt{2} }{2})^{2} \
I_{2}&= frac{I_{0}}{2} frac{1}{2}\
I_{2}&= frac{I_{0}}{4}
end{align*}
The light intensity that is emerging from the second polarizing filter [ framebox[1.1width]{$ therefore I_{2}= frac{I_{0}}{4} $}]

$$
I_{2}=frac{I_{0}}{4}
$$