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
1. 30 V/m
2. 6 V/m
3. 10 V/m
4. 24 V/m

Answer 1

Correct Answer - Option 4 : 24 V/m

Concept:

Energy of a light wave ∝ Intensity of the light wave

The intensity is given by the formula:

I = εvA²

Where, ‘ε’ is the permittivity of the medium in which light is travelling with velocity ‘v’ and ‘A’ is its amplitude.

Calculation:

Since, only 4% of the energy of the light gets reflected. Therefore, 96% of the energy of the light is transmitted.

E_transmitted (E_t) = 96% of E_incident (E_i)

\(\Rightarrow {{\rm{\varepsilon }}_0}{{\rm{\varepsilon }}_{\rm{r}}}{\rm{vA}}_{\rm{t}}^2 = \frac{{96}}{{100}} \times {{\rm{\varepsilon }}_0} \times {\rm{c}} \times {\rm{A}}_1^0\)

\(\Rightarrow A_t^2 = \frac{{96}}{{100}}\cdot\frac{{{\varepsilon _0}}}{{{\varepsilon _r}}}\cdot\frac{c}{v}A_i^2\)

\(\because \left[ {\sqrt {\frac{{{\varepsilon _0}}}{{{\varepsilon _r}}}} = \frac{v}{c}} \right]\)

\(\Rightarrow A_t^2 = \frac{{96}}{{100}}\cdot\frac{{{v^2}}}{{{c^2}}}\cdot\frac{c}{v}A_i^2\)

\(\Rightarrow A_t^2 = \frac{{96}}{{100}}\cdot\frac{v}{c}\cdot A_i^2\)

\(\because \left[ {\mu = \frac{c}{v} = 15} \right]\)

\(\Rightarrow A_t^2 = \frac{{96}}{{100}} \times \frac{1}{{3/2}} \times {30^2}\)

\(\Rightarrow {{\rm{A}}_{\rm{t}}} = \sqrt {\frac{{64}}{{100}} \times {{30}^2}} \)

∴ A_t = 24 V/m