A galaxy is moving away from the earth at a speed of 286 kms - 1. The shift in the wavelength of a redline at 630 nm is x \times 10 - 10 m. The value of x, to the nearest integer, is \underline\hspace2cm . [Take the value of speed of light c, as 3 \times 108 ms - 1]

Wave Optics PYQ — JEE Main 2021 · 18 Mar · Shift 2 · Q63 with Solution

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Correct answer: 6

Since the galaxy is moving away with speed $v \ll c$ , the wavelength shift is

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

where:

\frac{v}{c} = \frac{286 \times 10^3}{3 \times 10^8} = \frac{286}{3} \times 10^{-5} = 95.33 \times 10^{-5} = 9.533 \times 10^{-4}

\Delta \lambda = \lambda \cdot \frac{v}{c}

\Delta \lambda = 630 \times 10^{-9} \times 9.533 \times 10^{-4}

\Delta \lambda = (630 \times 9.533) \times 10^{-13}

\Delta \lambda \approx 6005.79 \times 10^{-13} = 6.00579 \times 10^{-10}\,\text{m}

So,

\Delta \lambda \approx 6 \times 10^{-10}\,\text{m}

Hence, comparing with $x \times 10^{-10}\,\text{m}$ ,

x \approx 6

To the nearest integer,

\boxed{6}

Stored correct answer = $6$

Our derived answer also = $6$ , so they agree.