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

A) $25\mathring{\mathrm{A}}$
B) Zero
C) $100\mathring{\mathrm{A}}$
D) $2.5\mathring{\mathrm{A}}$
Explanation

(a) $\mathrm{\Delta}\lambda = \lambda.\frac{v}{c} = \frac{1.5 \times 10^{6}}{3 \times 10^{8}} \times 5000 = 25\mathring{\mathrm{A}}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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