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Step-by-Step Solution
Step 1: Identify the given data
• Refractive index of glass, $n = 1.5$
• 4% of incident light intensity is reflected, so 96% is transmitted.
• Amplitude of the incident electric field, $A_i = 30\ \text{V/m}$.
• We need to find the amplitude of the transmitted electric field inside the glass, $A_t$.
Step 2: Relate intensities to amplitudes
The intensity $I$ of a wave (in a particular medium) is proportional to the square of its amplitude:
$I \propto A^2.$
When 4% of the incident intensity is reflected, 96% is transmitted. Hence,
$I_{t} = 0.96 \, I_{i}.$
Step 3: Account for medium dependence (impedance or refractive index)
The intensity in a medium can also be proportional to the factor involving its refractive index.
Typically, if $I \propto n A^2$, then for the same wave power, going from air (approx. $n=1$) to glass ($n=1.5$),
we adjust according to:
$ n_{\text{glass}}\,A_t^2 = \bigl(\text{fraction transmitted}\bigr) \times n_{\text{air}}\,A_i^2. $
Substituting $n_{\text{glass}}=1.5$ and $n_{\text{air}} \approx 1$, we get:
$ 1.5\,A_t^2 = 0.96 \times 1 \times A_i^2. $
Step 4: Compute the transmitted amplitude
From the above relation:
$ A_t^2 = \frac{0.96\,A_i^2}{1.5}. $
Given $A_i = 30$ V/m:
\[
A_t^2 = \frac{0.96 \times (30)^2}{1.5}
= \frac{0.96 \times 900}{1.5}
= 0.96 \times 600
= 576.
\]
Therefore,
\[
A_t = \sqrt{576} = 24\ \text{V/m}.
\]
Step 5: State the final answer
The amplitude of the electric field inside the glass slab is $24\ \text{V/m}$.
