Question:medium

How fast a person should drive his car so that the red signal of light appears green? (wavelengths of red and green colours are 6200 and 5400 respectively)

Show Hint

Wavelength shortening (\"blue shift\") occurs whenever an observer approaches a stationary light source. The ratio of the wavelength shift to the initial value gives the precise fraction of the speed of light at which the vehicle is traveling.
Updated On: Jun 7, 2026
  • \( 1.5\times10^{8}~\text{ms}^{-1} \)
  • \( 7\times10^{7}~\text{ms}^{-1} \)
  • \( 3.9\times10^{7}~\text{ms}^{-1} \)
  • \( 2\times10^{8}~\text{ms}^{-1} \)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Understand the colour change.
When you drive fast toward a light, its wavelength looks shorter (it shifts toward blue). Red light has a longer wavelength than green, so moving toward the red signal can make it look green.
Step 2: Write the Doppler shift for light.
For a source approaching, the change in wavelength is: \[ \Delta\lambda = \lambda\,\frac{v}{c} \] where $v$ is the car speed and $c = 3\times10^{8}\ ms^{-1}$.
Step 3: Find the needed wavelength change.
The red wavelength is $6200$ and the green is $5400$ (in angstrom units): \[ \Delta\lambda = 6200 - 5400 = 800 \]
Step 4: Rearrange for the speed.
\[ \frac{\Delta\lambda}{\lambda} = \frac{v}{c} \;\Rightarrow\; v = c\,\frac{\Delta\lambda}{\lambda} \]
Step 5: Put in the numbers.
Using the original red wavelength as $\lambda = 6200$: \[ v = (3\times10^{8})\frac{800}{6200} = (3\times10^{8})\frac{4}{31} \]
Step 6: Work out the value.
\[ v = \frac{12\times10^{8}}{31} \approx 3.9\times10^{7}\ ms^{-1} \] \[ \boxed{v \approx 3.9\times10^{7}\ ms^{-1}} \]
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