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(B) The rocket is moving away from the earth,so the observed wavelength will be redshifted (increased).Using the Doppler effect formula for light: $\f

Vedclass · Accepted Answer

$5520$

Vedclass · Answer

(B) The rocket is moving away from the earth,so the observed wavelength will be redshifted (increased).Using the Doppler effect formula for light: $\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$.Given: $v = 6 	imes 10^7 \ m/s$,$c = 3 	imes 10^8 \ m/s$,and $\lambda = 4600 \ \mathring A$.Calculating the fractional change: $\frac{\Delta \lambda}{\lambda} = \frac{6 	imes 10^7}{3 	imes 10^8} = 0.2$.Since the source is moving away,the observed wavelength $\lambda' = \lambda + \Delta \lambda = \lambda(1 + 0.2) = 1.2 \lambda$.$\lambda' = 1.2 	imes 4600 \ \mathring A = 5520 \ \mathring A$.

$A$ rocket is moving away from the earth at a speed of $6 \times 10^7 \ m/s$. The rocket has blue light in it. What will be the wavelength of light recorded by an observer on the earth in $\mathring A$? (Wavelength of blue light $\lambda = 4600 \ \mathring A$)

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