Q154 / 160 · TS EAMCET 2020 (10 Sep Shift 1) · Physics
What is the amplitude of the electric field in a parallel beam of light intensity $\left(\frac{15}{\pi}\right) \frac{\mathrm{W}}{\mathrm{m}^2}$ ? $\left[\right.$ Assume,$\left.\frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2}\right]$
- Correct answerA$60 \mathrm{~N} / \mathrm{C}$
- B$50 \mathrm{~N} / \mathrm{C}$
- C$40 \mathrm{~N} / \mathrm{C}$
- D$30 \mathrm{~N} / \mathrm{C}$
Solution
Intensity of parallel beam of light is given as
where, $E_0=$ amplitude of electric field. Here, $\begin{aligned} I & =\frac{15}{\pi} \frac{\mathrm{W}}{\mathrm{m}^2}, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2} \\ c & =3 \times 10^8 \mathrm{~ms}^{-1} \end{aligned}$ From Eq. (i), we get $\begin{aligned} & E_0^2=\frac{2 \times I}{\varepsilon_0 \times c}=\frac{2 \times 4 \pi \times I}{4 \pi \varepsilon_0 \times c}=\left(\frac{2 \times 4 \pi \times \frac{15}{\pi} \times 9 \times 10^9}{3 \times 10^8}\right) \\ \Rightarrow & E_0^2=3600 \Rightarrow E_0=60 \mathrm{~N} / \mathrm{C} \end{aligned}$
where, $E_0=$ amplitude of electric field. Here, $\begin{aligned} I & =\frac{15}{\pi} \frac{\mathrm{W}}{\mathrm{m}^2}, \frac{1}{4 \pi \varepsilon_0}=9 \times 10^9 \frac{\mathrm{Nm}^2}{\mathrm{C}^2} \\ c & =3 \times 10^8 \mathrm{~ms}^{-1} \end{aligned}$ From Eq. (i), we get $\begin{aligned} & E_0^2=\frac{2 \times I}{\varepsilon_0 \times c}=\frac{2 \times 4 \pi \times I}{4 \pi \varepsilon_0 \times c}=\left(\frac{2 \times 4 \pi \times \frac{15}{\pi} \times 9 \times 10^9}{3 \times 10^8}\right) \\ \Rightarrow & E_0^2=3600 \Rightarrow E_0=60 \mathrm{~N} / \mathrm{C} \end{aligned}$
Hence, the correct answer is option: 'A'.