91影视

The source wavelength,${位}_{source}=500\mathrm{nm}$

The observer wavelength,${位}_{obs}=500nm$

The velocity of the source, v = 0.8 c

The relationship between the source and observer wavelength is given as follows.

${\mathit{位}}_{\mathbf{s}\mathbf{o}\mathbf{u}\mathbf{r}\mathbf{c}\mathbf{e}}\mathbf{=}{\mathit{位}}_{\mathbf{o}\mathbf{b}\mathbf{s}}\frac{\sqrt{\mathbf{1}\mathbf{-}{\left({\displaystyle \frac{v}{c}}\right)}^{\mathbf{2}}}}{\mathbf{1}\mathbf{+}{\displaystyle \frac{\mathbf{v}}{\mathbf{c}}}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{胃}}$ 鈥︹�� (i)

Here, c is the speed of light, $\mathit{胃}$is the angle between the source of light and observer and v is the velocity of the source.

Calculate the angle between the source and observer light.

Substitute 500 nm for ${位}_{obs}\mathrm{and}{位}_{source}$into equation (i).

$$\begin{gathered} 500=500\frac{\sqrt{1-{\left({\displaystyle \frac{v}{c}}\right)}^2}}{1+{\displaystyle \frac{v}{c}}\cos 胃} \\ 1=\frac{\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{1+\frac{v}{c}\cos 胃} \\ 1+\frac{v}{c}\cos 胃=\sqrt{1-{\left(\frac{v}{c}\right)}^2} \\ \left(1+\frac{v}{c}\cos 胃\right)=1-{\left(\frac{v}{c}\right)}^2 \end{gathered}$$

Solve further as,

$$\begin{gathered} 1+{\left(\frac{v}{c}\right)}^2{\cos }^2胃+2\frac{v}{c}\cos 胃=1-{\left(\frac{v}{c}\right)}^2 \\ {\left(\frac{v}{c}\right)}^2{\cos }^2胃+2\frac{v}{c}\cos 胃+{\left(\frac{v}{c}\right)}^2=0 \end{gathered}$$

Solve the above quadratic equation,

$$\begin{gathered} \cos 胃=\frac{-2{\displaystyle \frac{v}{c}}卤\sqrt{{\left(2{\displaystyle \frac{v}{c}}\right)}^2-4{\left(\frac{v}{c}\right)}^2{\left(\frac{v}{c}\right)}^2}}{2{\left(\frac{v}{c}\right)}^2} \\ \cos 胃=\frac{-2{\displaystyle \frac{v}{c}}卤\sqrt{{\left(2{\displaystyle \frac{v}{c}}\right)}^2-4{\left(\frac{v}{c}\right)}^2}}{2{\left(\frac{v}{c}\right)}^2} \\ \cos 胃=\frac{-2{\displaystyle \frac{v}{c}}卤2{\left(\frac{v}{c}\right)}^2\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{2{\left(\frac{v}{c}\right)}^2} \\ \cos 胃=\frac{-2{\displaystyle \frac{v}{c}}卤\sqrt{1-{\left(\frac{v}{c}\right)}^2}}{v} \end{gathered}$$

Substitute 0.8c for v into above equation.

$$\begin{gathered} \cos 胃=\frac{-1卤\sqrt{1-{\left({\displaystyle \frac{0.8c}{c}}\right)}^2}}{\frac{0.8c}{c}} \\ \cos 胃=\frac{-1卤\sqrt{1-0.64}}{0.8} \\ \cos 胃=\frac{-1卤0.6}{0.8} \\ \cos 胃=-0.5\mathrm{or}-2 \end{gathered}$$

From the above solutions,$\cos 胃=-2$ is not valid solution. So, calculate the angle for -0.5 .

$$\begin{gathered} \cos 胃=-0.5 \\ 胃={\cos }^{-1}(-0.5) \\ 胃={120}^{掳} \end{gathered}$$

Since the angle is $120掳$that is other than $0掳$and $180掳$. Therefore, we can see the light from the beacon.