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A light wave is incident normally on a glass slab of refractive index 1.5 . If 4% of light gets refelcted 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 :
Option 1)
6 V/m
Option 2)
24 V/m
Option 3)
10 V/m
Option 4)
30 V/m

Answers (1)

best_answer

When Thermal Radiation(Q) fall in a body -

$Q=Q_{a}+Q_{t}+Q_{r}$

$\frac{Q}{Q}=\frac{Q_{a}}{Q}+\frac{Q_{t}}{Q}+\frac{Q_{r}}{r}$

1 = a + r + t

- wherein

$\mu =1.5$

For a wave

$I=\frac{1}{2}\varepsilon _{0}E_{0}^{2}c$

in medium

$I=\frac{1}{2}\varepsilon E^{2}V$

& $I^{2}=\left ( 1-0.04 \right ) I=0.96I$

$\Rightarrow \frac{1}{2}\varepsilon E^{2}V=0.96\times \frac{1}{2}\times \varepsilon _{0}E_{0}^{2}\times c$

$\Rightarrow E=\sqrt{0.96}\times \sqrt{\frac{\varepsilon _{0}}{\varepsilon }}\times \sqrt{\frac{C}{V}}\times E_{0}$

$\varepsilon =\varepsilon _{0}\varepsilon _{r}$ & $\mu =\sqrt{\mu _{r}\varepsilon _{r}},\mu =\frac{c}{v}$

So $E=\sqrt{0.96}\times \sqrt{\frac{\varepsilon _{0}}{\varepsilon_{0}\varepsilon _{r}}}\times \sqrt{\frac{C}{V}}\times E_{0}$

$=\sqrt{0.96}\times \frac{1}{\sqrt{\varepsilon _{r}}}\times \sqrt{\frac{C}{V}}\times E_{0}$

Let $\mu =1$

So $\mu =\sqrt{\varepsilon _{r}}$

So $E=\sqrt{0.96}\times \sqrt{\frac{1}{\varepsilon _{r}}}\times \sqrt{u}\times E_{0}$

$E=\sqrt{\frac{0.96}{\mu }} E_{0}=\sqrt{\frac{0.96}{1.5 }} E_{0}$

$E\approx 24\frac{V}{m}$

Option 1)

6 V/m

Option 2)
24 V/m
Option 3)

10 V/m

Option 4)

30 V/m

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