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Difficulty: Very hardElectromagnetic Waves and Electromagnetic Spectrum

An electromagnetic wave has a wavelength of 1.5×10−7 m1.5 \times 10^{-7}\text{ m} in a vacuum. It enters an optical medium where its speed decreases to 2.0×108 m/s2.0 \times 10^8\text{ m/s}. Given that the speed of light in vacuum is c=3.0×108 m/sc = 3.0 \times 10^8\text{ m/s}, what is the wavelength of the wave in the medium, and to which spectral region does its frequency belong?

  1. 1.0×10−7 m1.0 \times 10^{-7}\text{ m} and Ultraviolet regionAnswer
  2. B
    1.0×10−7 m1.0 \times 10^{-7}\text{ m} and Visible light region
  3. C
    2.25×10−7 m2.25 \times 10^{-7}\text{ m} and Ultraviolet region
  4. D
    1.5×10−7 m1.5 \times 10^{-7}\text{ m} and Infrared region

Answer

The wavelength in the medium is 1.0×10−7 m1.0 \times 10^{-7}\text{ m} and its frequency belongs to the Ultraviolet region.
When an electromagnetic wave transitions into a medium, its frequency ff remains constant at 2.0×1015 Hz2.0 \times 10^{15}\text{ Hz}, placing it strictly in the Ultraviolet spectrum. Its wavelength in the medium shrinks to λmed=vf=2.0×1082.0×1015=1.0×10−7 m\lambda_{med} = \frac{v}{f} = \frac{2.0 \times 10^8}{2.0 \times 10^{15}} = 1.0 \times 10^{-7}\text{ m}.

Step-by-Step Solution

1
Calculate the frequency of the wave in vacuum using the wave equation c=fλvacc = f \lambda_{vac}.
f=3.0×108 m/s1.5×10−7 m=2.0×1015 Hzf = \frac{3.0 \times 10^8\text{ m/s}}{1.5 \times 10^{-7}\text{ m}} = 2.0 \times 10^{15}\text{ Hz}.
Frequency is an intrinsic property determined by the source and does not change when entering a new medium.
2
Identify the electromagnetic spectrum region corresponding to the calculated frequency.
A frequency of 2.0×1015 Hz2.0 \times 10^{15}\text{ Hz} falls in the range 7.5×1014 Hz7.5 \times 10^{14}\text{ Hz} to 3.0×1016 Hz3.0 \times 10^{16}\text{ Hz}, which corresponds to the Ultraviolet region.
Spectral classifications are uniquely determined by frequency (or vacuum wavelength).
3
Determine the wavelength in the medium using λmed=vf\lambda_{med} = \frac{v}{f}.
λmed=2.0×108 m/s2.0×1015 Hz=1.0×10−7 m\lambda_{med} = \frac{2.0 \times 10^8\text{ m/s}}{2.0 \times 10^{15}\text{ Hz}} = 1.0 \times 10^{-7}\text{ m}.
The wavelength changes proportionally with phase velocity in a medium.

Key Concept

Invariance of electromagnetic wave frequency across media boundaries and wave equation relationship v=fλv = f \lambda.
Estimated Time:2m 0s
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Topics

Electromagnetic Waves and Electromagnetic Spectrum

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