Published by:
CGP EDU Academic Team
Published on: September 12, 2026
The earth is moving towards a fixed star with a velocity of 30 km s -1 . An observer on the earth observes a shift of 0.58 A in the wavelength of light coming from the star. What is the actual wavelength of light (in
m) emitted by the star?
Text Solution
Verified by ExpertsThe correct answer is:
A
To find the actual wavelength emitted by the star, we will use the formula for Doppler effect related to the shift in wavelength due to relative motion between the source and the observer.
Step 1: The observed shift in wavelength (Δλ) is given as 0.58 A. The velocity (v) of the observer (Earth) moving towards the source (star) is 30 km/s.
Step 2: The formula relating the observed shift in wavelength to the actual wavelength (λ) is given by:
$$ \Delta \lambda = \frac{\lambda_{actual} \cdot v}{c} $$
where c is the speed of light (approximately 3 x 10^8 m/s). Rearranging the formula gives us:
$$ \lambda_{actual} = \frac{\Delta \lambda \cdot c}{v} $$
Step 3: Substitute the known values into the equation:
$$ \lambda_{actual} = \frac{0.58 \times 10^{-10} \ m \cdot 3 \times 10^8 \ m/s}{30 \times 10^3 \ m/s} $$
Step 4: Calculate the actual wavelength:
$$ \lambda_{actual} = \frac{0.58 \times 10^{-10} \ m \cdot 3 \times 10^8}{30 \times 10^3} \approx 5.8 \times 10^{-7} \ m $$
Conclusion: Thus, the actual wavelength of the light emitted by the star is approximately 5.8 \times 10^{-7} m or 580 nm.
Step 1: The observed shift in wavelength (Δλ) is given as 0.58 A. The velocity (v) of the observer (Earth) moving towards the source (star) is 30 km/s.
Step 2: The formula relating the observed shift in wavelength to the actual wavelength (λ) is given by:
$$ \Delta \lambda = \frac{\lambda_{actual} \cdot v}{c} $$
where c is the speed of light (approximately 3 x 10^8 m/s). Rearranging the formula gives us:
$$ \lambda_{actual} = \frac{\Delta \lambda \cdot c}{v} $$
Step 3: Substitute the known values into the equation:
$$ \lambda_{actual} = \frac{0.58 \times 10^{-10} \ m \cdot 3 \times 10^8 \ m/s}{30 \times 10^3 \ m/s} $$
Step 4: Calculate the actual wavelength:
$$ \lambda_{actual} = \frac{0.58 \times 10^{-10} \ m \cdot 3 \times 10^8}{30 \times 10^3} \approx 5.8 \times 10^{-7} \ m $$
Conclusion: Thus, the actual wavelength of the light emitted by the star is approximately 5.8 \times 10^{-7} m or 580 nm.
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