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(A) The Doppler effect for light when the source is approaching the observer is given by the formula for the observed wavelength $\lambda^{\prime} = \

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$25$

Vedclass · Answer

(A) The Doppler effect for light when the source is approaching the observer is given by the formula for the observed wavelength $\lambda^{\prime} = \lambda \left(1 - \frac{v}{c}ight)$.Here,$\lambda = 5000 \; \mathring{A}$ is the original wavelength,$v = 1.5 	imes 10^6 \; m/s$ is the velocity of the star,and $c = 3 	imes 10^8 \; m/s$ is the speed of light.The change in wavelength $\Delta \lambda$ is given by $\Delta \lambda = \lambda - \lambda^{\prime}$.Substituting $\lambda^{\prime} = \lambda - \lambda \frac{v}{c}$,we get $\Delta \lambda = \lambda \frac{v}{c}$.Calculating the value: $\Delta \lambda = 5000 \; \mathring{A} 	imes \frac{1.5 	imes 10^6 \; m/s}{3 	imes 10^8 \; m/s}$.$\Delta \lambda = 5000 	imes 0.5 	imes 10^{-2} = 5000 	imes 0.005 = 25 \; \mathring{A}$.Thus,the change in wavelength is $25 \; \mathring{A}$.

$A$ star emitting radiation at a wavelength of $5000 \; \mathring{A}$ is approaching Earth with a velocity of $1.5 \times 10^6 \; m/s$. The change in wavelength of the radiation as received on Earth is ....... $\mathring{A}$.

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