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Step-by-Step Solution
Step 1: Recognize the Speed of Electromagnetic Wave in Vacuum and Medium
In free space (vacuum), the speed of an electromagnetic wave is
$ c = \frac{1}{\sqrt{\mu_0\,\epsilon_0}} $.
In a transparent (non-magnetic) medium of refractive index $n$, the speed of the electromagnetic wave becomes
$ v = \frac{1}{\sqrt{k\,\mu_0\,\epsilon_0}} $,
where $ k $ is the dielectric constant related to the refractive index by $ n = \sqrt{k} $.
Hence,
$ \frac{c}{v} = \sqrt{k} = n. $
Step 2: Relate Intensity with Electric Field
The intensity $ I $ of an electromagnetic wave in free space can be written as
$ I = \frac{1}{2}\,\epsilon_0\,E_0^2\,c, $
where $ E_0 $ is the amplitude of the electric field in vacuum.
In a medium (with refractive index $ n $), the intensity is
$ I' = \frac{1}{2}\,(\epsilon_0 k)\,E^2\,v, $
where $ E $ is the amplitude of the electric field within the medium and $ v $ is the speed of the wave in that medium.
Step 3: Derive the Ratio of Electric Fields
Since the wave enters the medium without loss, the intensity before and after remains the same numerically (50 W m–2 in this case, but the exact value is not needed to find the ratio). Equating the expressions for intensity in vacuum and in the medium (and simplifying), we get:
$ \frac{1}{2}\,\epsilon_0\,E_0^2\,c \;=\; \frac{1}{2}\,\epsilon_0\,k\,E^2\,v. $
Cancel out the common factors and use $ k = n^2 $ and $ \frac{v}{c} = \frac{1}{n} $, to obtain:
$ E_0^2 \,c = k\,E^2\,v \quad\Rightarrow\quad \frac{E_0^2}{E^2} = \frac{k\,v}{c}
= \frac{n^2}{n} = n. $
Taking the square root of both sides:
$ \frac{E_0}{E} = \sqrt{n} \quad\Longrightarrow\quad \frac{E}{E_0} = \frac{1}{\sqrt{n}}. $
Thus, the electric field inside the medium is smaller by a factor of $ \sqrt{n} $ than in vacuum, or equivalently
$ E_0 : E = \sqrt{n} : 1. $
Step 4: Derive the Ratio of Magnetic Fields
Similarly, the magnetic field intensities in free space ($ B_0 $) and in the medium ($ B $) can be related through the intensity or from the wave speed relation. A similar argument shows:
$ \frac{B_0}{B} = \frac{1}{\sqrt{n}} \quad\Longrightarrow\quad \frac{B}{B_0} = \sqrt{n}. $
Step 5: Conclude the Ratios
Putting these together, the ratio of the magnitudes of electric fields (vacuum to medium) and magnetic fields (vacuum to medium) is:
$ \left(\frac{E_0}{E},\,\frac{B_0}{B}\right) = \left(\sqrt{n},\,\frac{1}{\sqrt{n}}\right). $
Hence, the correct answer is
$ \bigl(\sqrt{n},\,\tfrac{1}{\sqrt{n}}\bigr). $
