An observer moves with a speed of 0.86 C towards a stationary source of light in air. The speed of light would appear to the observer to be:

Correct option is A

​$$\textbf{Concept:} \\\text{In special relativity, the speed of light in vacuum (or air) is always constant regardless of the motion of the source or observer.} \\c = 3 \times 10^8\ \text{m/s} \\\text{Hence, the speed of light remains unchanged for all observers.} \\[10pt]\textbf{Analysis Using Doppler Shift (for frequency \& wavelength):} \\\bullet\ \text{Source speed } u_s = 0 \\\bullet\ \text{Observer speed } u_o = 0.86c \\[6pt]\text{The apparent frequency:} \\f' = \left( \frac{v + u_o}{v} \right) f_0 = \left( \frac{c + 0.86c}{c} \right) f_0 = 1.86 f_0 \\[6pt]\text{The actual wavelength:} \\\lambda = \frac{c}{f_0} \\[6pt]\text{The apparent wavelength:} \\\lambda' = \frac{c}{f'} = \frac{c}{1.86 f_0} = \frac{\lambda}{1.86} \\[6pt]\text{Now, speed of light as seen by observer:} \\v = \lambda' \cdot f' = \frac{\lambda}{1.86} \cdot 1.86 f_0 = \lambda f_0 = c$$​​