Question

Speed of Light in vacuum is :

• A. $3 \times 10^6$ m/s
• B. $3 \times 10^8$ m/s
• C. $3 \times 10^5$ m/s
• D. $3 \times 10^7$ m/s

Hint:

A common mistake is choosing $3 \times 10^5 \text{ km/s}$. While that value is correct in kilometers, always check the units. In meters per second (m/s), which is the SI standard, the exponent must be $10^8$.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question asks for the standard value of the speed of light in a vacuum, a universal constant in physics denoted by the symbol \( c \).
Step 2: Detailed Explanation:

• Exact Value: The speed of light in a vacuum is defined with a precise value of $299,792,458$ meters per second. For most scientific and engineering calculations, this is conveniently approximated to $3 \times 10^8 \text{ m/s}$.
  
• Universal Constant: A fundamental postulate of Einstein's Theory of Special Relativity is that the speed of light in a vacuum is the same for all observers, irrespective of the motion of the light source. It represents the maximum speed at which energy, matter, and information can travel.
  
• Significance in Calculations: This constant is a key component in numerous famous physics equations, including the mass-energy equivalence formula, \( E = mc^2 \). It is also used to define astronomical distances, such as the light-year.
  
• Speed in Media: Light slows down when it passes through a transparent medium like air, water, or glass. The factor by which it slows is known as the medium's "refractive index" (\( n = c/v \)).
  
• Magnitude Comparison: To put its speed into perspective, a beam of light could travel around the Earth's equator about 7.5 times in just one second. This immense speed is why light travel appears instantaneous in our everyday experience.
  

Step 3: Final Answer:
The widely accepted scientific value for the speed of light in a vacuum is $3 \times 10^8 \text{ m/s}$.

Answer 2

Instead of simply quoting the constant, its value can be checked using the wave relationship \( c = f\lambda \), where \(f\) is frequency and \(\lambda\) is wavelength, applied to visible light, whose frequency and wavelength are independently well known.

1. \(3 \times 10^6\) m/s: Substituting a typical visible-light frequency of about \(5 \times 10^{14}\text{ Hz}\) and wavelength of about \(6 \times 10^{-7}\text{ m}\) gives \(c = (5\times10^{14})(6\times10^{-7}) \approx 3\times10^{8}\text{ m/s}\), which is three orders of magnitude larger than this option, so it is far too small.
2. \(3 \times 10^8\) m/s: This matches the value obtained directly from the frequency-wavelength calculation above, and is also the value that keeps orbital light-travel times (like roughly 8 minutes from the Sun to Earth, a distance of about \(1.5\times10^{11}\text{ m}\)) consistent: \( t = d/c \approx (1.5\times10^{11})/(3\times10^8) \approx 500\text{ s} \approx 8.3 \) minutes, matching observation.
3. \(3 \times 10^5\) m/s: This is a thousand times smaller than the correct value, roughly the speed of a very fast artillery shell relative to light, nowhere close to an electromagnetic wave's speed.
4. \(3 \times 10^7\) m/s: Using this value in the Sun-Earth light travel time calculation would give a travel time ten times too long, contradicting the well-established roughly 8-minute figure, so it can't be correct.

Both the wave-relation calculation and the Sun-to-Earth timing check converge on the same magnitude.

Therefore, the correct answer is \(3 \times 10^8\) m/s.