$$
tt{The Direction of the electric field of a electromagnetic wave represent its polarization in general, Therefore ins this example the light is horizontally polarized }
$$

$$
tt{(B)}
$$

$$
tt{Reffering to {color{#4257b2}{Figure 15.8 }}, we can see that electromagnetic waves are divided into groups based on their frequency and wavelength the visible portion of the spectrum sits in the 4.29 * 1014 Hz to 7.50 * 1014 interval, Looking into this portion we can notice that the colors perceived change based on the exact value of the frequency (can also be determined by the wavelength corresponding to that frequency.) }
$$

$$
tt{(C)}
$$

$$
tt{According to {color{#4257b2}{Figure 15.20}}, if we substract yellow light from an object emitting blue light, the human eye will perceive the color black}
$$

$$
tt{(B)}
$$

$$
tt{According to figure {color{#4257b2}{Figure 15.8}}, the correct order of electromagnetic wave from lowest to highest frequency is infrared, blue, ultraviolet, x-ray }
$$

$$
tt{(B)}
$$

$$
tt{Reffering to {color{#4257b2}{Figure 15.25}}, we see that the incident light comes in all direction of polarization, due to the unique transmission axis direction of the polarizer, all of theses incident light beams get absorbed partially based on their relative angle to the axis, summing all of these beams intensities before and after getting through the polarizer we can say that $I_f=frac{1}{2}I_i$ }
$$

$$
tt{(A)}
$$

$$
tt{Every electromagnetic obeys to the following formula in vacuum: $c=lambda f (f=frac{c}{lambda}) $ on the other hand the doppler formula states the following; }
$$

$$
f_{observed}=f_{source}(1pmfrac{v_{source}}{c})
$$

$$
frac{f_{observed}}{f_{source}}=(1pmfrac{v_{source}}{c}) Rightarrow frac{lambda_{source}}{lambda_{observed}}=(1pmfrac{v_{source}}{c})
$$

$$
frac{lambda_{source}}{lambda_{observed}}=frac{470.8}{537.4}=0.876<1
$$

$$
1pmfrac{v_{source}}{c}<1 Rightarrow pmfrac{v_{source}}{c}<0
$$

Therefore the final version of the doppler formula applied to this situation is $f_{observed}=f_{source}(1-frac{v_{source}}{c})$, According to the doppler formula the star is $text{color{#4257b2}{moving away}}$ from earth with a speed of:

$$
v_{source}=(1-frac{lambda_{source}}{lambda_{observed}})c=(1-frac{470.8}{537.4})c=boxed{0.12c}
$$

$$
tt{(B)}
$$

$$
tt{Every electromagnetic wave obeys to the following formula $c=lambda f (f=frac{c}{lambda}$ so if$ lambdain[400nm-700nm]$ then $f$:}
$$

$$
f_1=frac{c}{lambda_1}=frac{3*10^8}{400*10^{-9}}=7.5*10^{14}text{Hz}
$$

$$
f_2=frac{c}{lambda_2}=frac{3*10^8}{700*10^{-9}}=4.3*10^{14}text{Hz}
$$

$$
fin[4.3*10^{14}text{Hz}-7.5*10^{14}text{Hz}]
$$

$$
tt{(C)}
$$