Ncert Exercises & Solutions

The electric field produced by the radiations coming from \(100~\text{W}\) bulb at a \(3~\text{m}\) distance is \(E\). The electric field intensity produced by the radiations coming from \(50~\text{W}\) bulb at the same distance is:
1. \(\dfrac{E}{2}\)
2. \(2E\)
3. \(\dfrac{E}{\sqrt2}\)
4. \(\sqrt2E\)

Hint: \(P\propto E^2\)

Step: Find the electric field intensity produced by radiation from \(50~\text{W}\).
The intensity produced by the bulb is given by;
\(I = \frac{P}{4\pi r^2}~~~...(1)\)
The electric field intensity produced by radiation is given by;
\(I = \frac{1}{2}\varepsilon_0E^2c~~~...(2)\)
From equations \((1)\) and \((2)\) we get;
\(P \propto E^2\)
\(\Rightarrow E \propto \sqrt{P}\)
\(\Rightarrow \frac{E_2}{E_1}= \sqrt{\frac{P_2}{P_1}}\)
\(\Rightarrow \frac{E_2}{E}= \sqrt{\frac{50}{100}}=\sqrt{\frac{1}{2}}\)
\(\Rightarrow E_2 = \frac{E}{\sqrt{2}}\)
Hence, option (3) is the correct answer.

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