Light of wavelength 550 nm in vacuum is found to travel at 1.96 × 108 m/s in a certain liquid. Determine the index of refraction of the liquid.

Question

BlackTom AI · Accepted Answer

Index of refraction n = c / v.

1) Relevant concepts: The refractive index n of a medium is the ratio of the speed of light in vacuum c to the speed of light in the medium v: n = c/v. The given vacuum wavelength (550 nm) is not needed to find n but would let you compute the wavelength inside the liquid via λ_medium = λ_vacuum / n.

2) Substitute numerical values: take c = 3.00 × 10^8 m/s and v = 1.96 × 10^8 m/s.

3) Calculation (showing intermediate steps):
n = c / v = (3.00 × 10^8 m/s) / (1.96 × 10^8 m/s)
The 10^8 factors cancel, leaving n = 3.00 / 1.96.
Compute the division: 3.00 ÷ 1.96 = 1.5306122449… ≈ 1.53 (to three significant figures).

4) Why this is correct: The formula n = c/v directly gives how much slower light travels in the medium compared with vacuum. A value of n ≈ 1.53 means light travels about 1/1.53 ≈ 0.653 of its vacuum speed, consistent with v = 1.96×10^8 m/s. Refractive indices for ordinary transparent liquids commonly lie around 1.3–1.6, so 1.53 is physically reasonable.

5) Why the other choices are incorrect:
- 0.65 would imply v > c (since n < 1), which is not physically allowed for light propagation in passive media.
- 1.96 might confuse the numeric value of v (1.96×10^8 m/s) with n; it does not satisfy n = c/v.
- 5.50 is far too large and does not result from c/v with the given v (it appears to confuse the given wavelength 550 nm with n).

Conclusion: The correct index of refraction is 1.53, matching the provided answer.

Light of wavelength 550 nm in vacuum is found to travel at 1.96 × 108 m/s in a certain liquid. Determine the index of refraction of the liquid.单项选择题

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