### Waves - Result Question 79

#### 81. 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

(a) $0.25 \AA$

(b) $2.5 \AA$

(c) $25 \AA$

(d) $250 \AA$

[1996]

## Show Answer

#### Answer:

Correct Answer: 81. (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]