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(C) According to the relativistic Doppler effect,when a source approaches the observer,the observed wavelength $\lambda'$ is given by $\lambda' = \lam

Vedclass · Accepted Answer

$3167$

Vedclass · Answer

(C) According to the relativistic Doppler effect,when a source approaches the observer,the observed wavelength $\lambda'$ is given by $\lambda' = \lambda \sqrt{\frac{1 - v/c}{1 + v/c}}$.Given $\lambda = 5500 \ \mathring{A}$ and $v = 0.8 c$.Substituting the values: $\lambda' = 5500 \sqrt{\frac{1 - 0.8}{1 + 0.8}} = 5500 \sqrt{\frac{0.2}{1.8}} = 5500 \sqrt{\frac{1}{9}} = 5500 	imes \frac{1}{3} \approx 1833.3 \ \mathring{A}$.The Doppler shift is the difference between the original wavelength and the observed wavelength: $\Delta\lambda = \lambda - \lambda' = 5500 - 1833.3 = 3166.7 \ \mathring{A} \approx 3167 \ \mathring{A}$.

$A$ light source approaches the observer with velocity $0.8 c$. The Doppler shift for the light of wavelength $5500 \ \mathring{A}$ is.....$\mathring{A}$.

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