Waves Ques 79
79. A star, which is emitting radiation at a wavelength of $5000 $ $\AA$, is approaching the earth with a velocity of $1.50 \times 10^{6}$ $ m / s$. The change in wavelength of the radiation as received on the earth is
[1996]
(a) $0.25 $ $\AA$
(b) $2.5 $ $\AA$
(c) $25 $ $\AA$
(d) $250 $ $\AA$
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Answer:
Correct Answer: 79.(c)
Solution:
- (c) Given : Wavelength $(\lambda)=5000$ $ \AA$ velocity of star $(v)=1.5 \times 10^{6}$ $ m / s$.
We know that wavelength of the approaching $star(\lambda^{\prime})=\lambda(\frac{c-v}{c})$
or, $\frac{\lambda^{\prime}}{\lambda}=\frac{c-v}{c}=1-\frac{v}{c}$
or, $\frac{v}{c}=1-\frac{\lambda^{\prime}}{\lambda}=\frac{\lambda-\lambda^{\prime}}{\lambda}=\frac{\Delta \lambda}{\lambda}$
Therefore, $\Delta \lambda=\lambda \times \frac{v}{c}=5000 \times \frac{1.5 \times 10^{6}}{3 \times 10^{8}}=25$ $ \AA$
[where $\Delta \lambda=$ Change in the wavelength]
BETA
