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
- A \(10.5 \mathrm{~W} \mathrm{~m}^{-2}\)
- B \(5.25 \mathrm{~W} \mathrm{~m}^{-2}\)
- C \(3.5 \mathrm{~W} \mathrm{~m}^{-2}\)
- D \(1.75 \mathrm{~W} \mathrm{~m}^{-2}\)
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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