Optics 7 Question 3
3. 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 propogating in the glass medium will be
(2019 Main, 12 Jan I)
(a) $30 V / m$
(b) $6 V / m$
(c) $10 V / m$
(d) $24 V / m$
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Solution:
- Energy of a light wave $\propto$ Intensity of the light wave
Since, intensity $=\varepsilon v A^{2}$
where, $\varepsilon$ is the permittivity of the medium in which light is travelling with velocity $v$ and $A$ is its amplitude.
Since, only $4 %$ of the energy of the light gets reflected.
$\therefore 96 %$ of the energy of the light is transmitted.
$$ \begin{array}{rlrl} E _{\text {transmitted }}\left(E _t\right) & =96 % \text { of } E _{\text {incident }}\left(E _i\right) & \\ \varepsilon _0 \varepsilon _r v A _t^{2} & =\frac{96}{100} \times \varepsilon _0 \times c \times A _1^{0} & \\ A _t^{2} & =\frac{96}{100} \cdot \frac{\varepsilon _0}{\varepsilon _r} \cdot \frac{c}{v} A _i^{2} & \\ A _t^{2} & =\frac{96}{100} \cdot \frac{v^{2}}{c^{2}} \cdot \frac{c}{v} A _i^{2} & \because \sqrt{\frac{\varepsilon _0}{\varepsilon _r}}=\frac{v}{c} \\ A _t^{2} & =\frac{96}{100} \cdot \frac{v}{c} \cdot A _i^{2} & \\ & =\frac{96}{100} \times \frac{1}{3 / 2} \times 30^{2} & \because \mu=\frac{c}{v}=1.5 \\ A _t & =\sqrt{\frac{64}{100} \times 30^{2}} \\ A _t & =24 V / m \end{array} $$
