Question:
A light wave is incident normally on a glass slab of refractive index 1.5. If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be :
A light wave is incident normally on a glass slab of refractive index 1.5. If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V/m, then the amplitude of the electric field for the wave propagating in the glass medium will be :
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Intensity in a medium is proportional to \(nE_0^2\); the transmitted intensity is 96% of the incident intensity.
Updated On: Nov 5, 2026
- \(6\ \text{V/m}\)
- \(10\ \text{V/m}\)
- \(30\ \text{V/m}\)
- \(24\ \text{V/m}\)
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The Correct Option is D
Solution and Explanation
Intensity \(I = \frac{1}{2}\varepsilon_0 c\, n E_0^2\), so \(I \propto nE_0^2\) in a medium of index n.
Reflected 4% means transmitted intensity \(I_t = 0.96\, I_i\).
\(n\,E_t^2 = 0.96\,(1)\,E_i^2 \Rightarrow 1.5\,E_t^2 = 0.96\times 900 = 864\).
\(E_t^2 = 576 \Rightarrow E_t = 24\ \text{V/m}\).
Answer: (4) 24 V/m
Reflected 4% means transmitted intensity \(I_t = 0.96\, I_i\).
\(n\,E_t^2 = 0.96\,(1)\,E_i^2 \Rightarrow 1.5\,E_t^2 = 0.96\times 900 = 864\).
\(E_t^2 = 576 \Rightarrow E_t = 24\ \text{V/m}\).
Answer: (4) 24 V/m
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