Optics Question 252

Question: A light source approaches the observer with velocity 0.8 c. The doppler shift for the light of wavelength 5500AA is

[MP PET 1996]

Options:

A) 4400AA

B) 1833AA

C) 3167AA

D) 7333AA

Show Answer

Answer:

Correct Answer: C

Solution:

According to Doppler’s principle λ=λ1v/c1+v/c

for v = c λ=5500(10.8)1+0.8=1833.3

Shift =55001833.3=3167AA



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