Question

A spaceship approaches Earth at a speed of 0.42c. A light on the front of the ship appears red (wavelength 650 nm) to passengers on the ship. What (a) wavelength and (b) color (blue, green, or yellow) would it appear to an observer on Earth?

Short answer

(a) The wavelength of light for observer on Earth is $415nm.$

(b) The color of light for observer on Earth is blue.

Step-by-step solution

Identification of given data

The wavelength of light is ${\lambda }_o=650nm.$

The speed of spaceship is $v=0.42c$

The wavelength of light for observer on Earth is calculated by using the formula for Doppler effect in light and color of light is decided by wavelength of light.

Determination of wavelength of light for observer on Earth

(a)

The wavelength of light for observer on Earth is given as

$\lambda ={\lambda }_o\sqrt{\left(\frac{1+{\displaystyle \frac{v}{c}}}{1-{\displaystyle \frac{v}{c}}}\right)}$

Here, $c$ is the speed of light and its value is $3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Substitute all the values in the above equation.

$$\begin{gathered} \lambda =\left(615mm\right)\sqrt{\frac{\left(1+{\displaystyle \frac{0.42c}{c}}\right)}{\left(1-{\displaystyle \frac{0.42c}{c}}\right)}} \\ \lambda =415nm \end{gathered}$$

Therefore, the wavelength of light for observer on Earth is $415nm.$

$\lambda ={\lambda }_o\sqrt{\left(\frac{1+{\displaystyle \frac{v}{c}}}{1-{\displaystyle \frac{v}{c}}}\right)}$

Identification for color by wavelength

(b)

The wavelength of the light falls between wavelength of blue and green on visible spectrum but it is nearer to blue color wavelength so color of light for observer on Earth is blue.

Therefore, the colour of light for observer on Earth is blue.