An electromagnetic wave of electric‑field amplitude $E_{0} = 200 V m^{- 1}$ is incident normally on a glass plate (refractive index $n = 1.5$ ) from air ( $n = 1$ ). Find the magnitude of the reflected electric‑field amplitude.

medium 2 marks Class 12 · Physics · Electromagnetic Waves

Answer

40\

V m^{- 1}

The correct answer is 40\ $V m^{- 1}$ . — Present this value as the answer.

Step-by-step solution

For normal incidence, the amplitude reflection coefficient for the electric field is $r = \frac{n _{2} - n _{1}}{n _{2} + n _{1}}$ .
Identify $n_{1} = 1$ (air) and $n_{2} = 1.5$ (glass).
Calculate $r = \frac{1.5 - 1}{1.5 + 1} = \frac{0.5}{2.5} = 0.2$ .
The reflected amplitude is $E_{r} = r E_{0} = 0.2 \times 200 V m^{- 1} = 40 V m^{- 1}$ .
Present this value as the answer.

Common mistake: Using intensity (proportional to

E^{2}

) instead of amplitude when applying the reflection coefficient.

Concept tested: Reflection coefficient at normal incidence

Published 8 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked