Propagation In A Medium
A transparent dielectric medium is characterised by a relative permittivity of 4 and a relative permeability of 1. With what speed does an electromagnetic wave propagate through this medium?
Select the correct option:
Solution
1.5×108 m/s
Inside a material medium an electromagnetic wave slows because the medium's permittivity and permeability replace those of free space, giving v = \frac{1}{\sqrt{\mu \varepsilon}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}. With \varepsilon_r = 4 and \mu_r = 1, the denominator becomes \sqrt{4 \times 1} = 2, so v = (3.0 \times 10^{8})/2 = 1.5 \times 10^{8}\ \text{m/s}. The 3.0 \times 10^{8}\ \text{m/s} option ignores the medium and quotes the vacuum speed. The 0.75 \times 10^{8}\ \text{m/s} value wrongly divides by \varepsilon_r itself rather than its square root, giving a factor of four. The 6.0 \times 10^{8}\ \text{m/s} option, faster than light, is unphysical and would require multiplying instead of dividing. The effective refractive index here is n = \sqrt{\mu_r \varepsilon_r} = 2, exactly the factor by which the speed drops, in line with the NCERT and JEE treatment linking optics to electromagnetism. While the wave slows, its frequency stays fixed by the source, so it is the wavelength inside the medium that shrinks by the same factor of two. This reduction in wavelength while frequency is preserved is the microscopic origin of refraction observed in ordinary optics. As a plausibility check, any genuine medium must give a speed below c, and 1.5 \times 10^{8}\ \text{m/s} satisfies that requirement.
🔒 Solution Hidden from View
Submit your answer to unlock the detailed step-by-step solution.
About This Question
- Subject
- physics
- Chapter
- electromagnetic waves
- Topic
- propagation in a medium
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
1.5×108 m/s
Inside a material medium an electromagnetic wave slows because the medium's permittivity and permeability replace those of free space, giving v = \frac{1}{\sqrt{\mu \varepsilon}} = \frac{c}{\sqrt{\mu_r \varepsilon_r}}. With \varepsilon_r = 4 and \mu_r = 1, the denominator becomes \sqrt{4 \times 1} = 2, so v = (3.0 \times 10^{8})/2 = 1.5 \times 10^{8}\ \text{m/s}. The 3.0 \times 10^{8}\ \text{m/s} option ignores the medium and quotes the vacuum speed. The 0.75 \times 10^{8}\ \text{m/s} value wrongly divides by \varepsilon_r itself rather than its square root, giving a factor of four. The 6.0 \times 10^{8}\ \text{m/s} option, faster than light, is unphysical and would require multiplying instead of dividing. The effective refractive index here is n = \sqrt{\mu_r \varepsilon_r} = 2, exactly the factor by which the speed drops, in line with the NCERT and JEE treatment linking optics to electromagnetism. While the wave slows, its frequency stays fixed by the source, so it is the wavelength inside the medium that shrinks by the same factor of two. This reduction in wavelength while frequency is preserved is the microscopic origin of refraction observed in ordinary optics. As a plausibility check, any genuine medium must give a speed below c, and 1.5 \times 10^{8}\ \text{m/s} satisfies that requirement.
This medium difficulty physics question is from the chapter electromagnetic waves, covering the topic of propagation in a medium. It appeared in the 2025 exam.
Looking for more practice? Explore all physics questions or browse electromagnetic waves questions on RankGuru.