Physics · Electromagnetic Waves

JEE Main 2025 — 4 April, Evening Shift — Question 63

If an optical medium possesses a relative permeability of 10π\frac{10}{\pi} and relative permittivity of

10.0885\frac{1}{0.0885}, then the velocity of light is greater in vacuum than that in this medium by

\qquad times. (μ0=4π×10−7H/m,∈0=8.85×10−12 F/m\left(\mu_{0}=4 \pi \times 10^{-7} \mathrm{H} / \mathrm{m}, \in_{0}=8.85 \times 10^{-12} \mathrm{~F} / \mathrm{m}\right.,

c=3×108 m/sc=3 \times 10^{8} \mathrm{~m} / \mathrm{s} )

Answer: 6

Numerical answer — enter this value.

Step-by-step solution

μr=10πϵr=10.0885\mu_{r}=\frac{10}{\pi} \quad \epsilon_{r}=\frac{1}{0.0885}

c=1μ0∈0c=\frac{1}{\sqrt{\mu_{0} \in_{0}}}

v=1μrμ0∈r∈0=1μr∈rcv=\frac{1}{\sqrt{\mu_{r} \mu_{0} \in_{r} \in_{0}}}=\frac{1}{\sqrt{\mu_{r} \in_{r}}} c

v≃C6v \simeq \frac{C}{6} c≃6vc \simeq 6 v

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Subject
Physics
Chapter
Electromagnetic Waves
Topic
Properties of EM Waves and Electromagnetic Spectrum