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AP EAMCET · PHYSICS · Wave Optics

If \(11 \%\) of the power of a 200 W bulb is converted to visible radiation, then the intensity of the light at a distance of 100 cm from the bulb is

  1. A \(10.5 \mathrm{~W} \mathrm{~m}^{-2}\)
  2. B \(5.25 \mathrm{~W} \mathrm{~m}^{-2}\)
  3. C \(3.5 \mathrm{~W} \mathrm{~m}^{-2}\)
  4. D \(1.75 \mathrm{~W} \mathrm{~m}^{-2}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(1.75 \mathrm{~W} \mathrm{~m}^{-2}\)

Step-by-step Solution

Detailed explanation

\(P_{visible} = 200 \mathrm{~W} \times 0.11 = 22 \mathrm{~W}\) \(r = 100 \mathrm{~cm} = 1 \mathrm{~m}\) \(I = \frac{P_{visible}}{4\pi r^2} = \frac{22 \mathrm{~W}}{4\pi (1 \mathrm{~m})^2}\)…
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