A plane electromagnetic wave of electric field amplitude $E_{0} = 1.0 \times 1 0^{5} V/m$ is incident normally on a glass plate with refractive index $n_{2} = 1.5$ from air ( $n_{1} = 1$ ). Find the magnitude of the reflected electric field amplitude.

medium 3 marks Class 12 · Physics · Electromagnetic Waves

Answer

2.0 \times 1 0^{4} V/m

The correct answer is $2.0 \times 1 0^{4} V/m$ — Thus the reflected electric field amplitude has a magnitude of $2.0 \times 1 0^{4} V/m$ .

Step-by-step solution

The Fresnel reflection coefficient for normal incidence from medium 1 to medium 2 is $r = \frac{n _{1} - n _{2}}{n _{1} + n _{2}}$ .
Substitute $n_{1} = 1$ and $n_{2} = 1.5$ : $r = \frac{1 - 1.5}{1 + 1.5} = \frac{- 0.5}{2.5} = - 0.20$ (the negative sign indicates a phase reversal).
The reflected electric field amplitude is $E_{r} = r E_{0}$ .
Multiply the amplitude: $E_{r} = (- 0.20) (1.0 \times 1 0^{5} V/m) = - 2.0 \times 1 0^{4} V/m$ .
The magnitude of the reflected field is $∣ E_{r} ∣ = 2.0 \times 1 0^{4} V/m$ .
Thus the reflected electric field amplitude has a magnitude of $2.0 \times 1 0^{4} V/m$ .

Common mistake: ignoring negative sign or using wrong formula for reflection coefficient

Concept tested: Fresnel reflection at normal incidence