For a specific wavelength 670 nm of light coming from a galaxy moving with velocity v, the observed wavelength is 670.7 nm. The value of v is :

Wave Optics PYQ — JEE Main 2022 · 26 Jun · Shift 2 · Q62 with Solution

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Correct answer: C

Given data
- Emitted wavelength: $\lambda = 670\,\text{nm}$
- Observed wavelength: $\lambda' = 670.7\,\text{nm}$
Since the observed wavelength is larger, there is a red shift, so the galaxy is moving away.
Use Doppler shift for light (small velocity approximation)

For $v \ll c$ ,
$\frac{\Delta \lambda}{\lambda} = \frac{v}{c}$
where
$\Delta \lambda = \lambda' - \lambda = 670.7 - 670 = 0.7\,\text{nm}$
Substitute values
$\frac{v}{c} = \frac{0.7}{670}$ $v = c\left(\frac{0.7}{670}\right)$
Taking $c = 3 \times 10^8\,\text{m/s}$ ,
$v = 3 \times 10^8 \times \frac{0.7}{670}$
Calculate
$\frac{0.7}{670} \approx 1.0448 \times 10^{-3}$
Hence,
$v \approx 3 \times 10^8 \times 1.0448 \times 10^{-3}$ $v \approx 3.13 \times 10^5\,\text{m/s}$
Check options
- A: $3 \times 10^8\,\text{m/s}$ ❌
- B: $3 \times 10^{10}\,\text{m/s}$ ❌
- C: $3.13 \times 10^5\,\text{m/s}$ ✅
- D: $4.48 \times 10^5\,\text{m/s}$ ❌
Final answer

The correct option is
$\boxed{\text{C }\; 3.13 \times 10^5\,\text{m/s}}$