Optics Ques 204
- 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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Answer:
Correct Answer: 204.(d)
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.
$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} \quad [\because \sqrt{\frac{\varepsilon _0}{\varepsilon _r}}=\frac{v}{c}] $
$A _t^{2} =\frac{96}{100} \cdot \frac{v}{c} \cdot A _i^{2} $
$\quad =\frac{96}{100} \times \frac{1}{3 / 2} \times 30^{2} \quad [\because \mu=\frac{c}{v}=1.5] $
$ A _t =\sqrt{\frac{64}{100} \times 30^{2}} $
$A _t =24$ $ V / m$
BETA
